Q.Evaluate the definite integral: $\int_{0}^{\pi/4} (2 \sec^2 x + x^3 + 2) \, dx$

Showing the 12 most recent of 17 on this concept.

1. TG EAPCET 2026Set eng-2026-05-10-AN1 markMCQ
   Q.$\int e^{\tan x} \left( \tan^7 x + 5 \tan^6 x + \tan^5 x + 5 \tan^4 x \right) dx =$ (A) $e^{\tan x} \left( \dfrac{\tan^8 x}{8} \right) + c$ (B) $e^{\tan x} \left( \tan^5 x \right) + c$ (C) $e^{\tan x} \left( \dfrac{\tan^6 x}{6} \right) + c$ (D) $e^{\tan x} \left( \tan^7 x \right) + c$

   ›Reveal solutionSolution

   The integrand is exactly $\dfrac{d}{dx}\big(e^{\tan x}\tan^5 x\big)$, so the integral is $e^{\tan x}\tan^5 x + c$.

   Differentiate the candidate $e^{\tan x}\tan^5 x$:

   $\frac{d}{dx}\big(e^{\tan x}\tan^5 x\big) = e^{\tan x}\sec^2 x\cdot \tan^5 x + e^{\tan x}\cdot 5\tan^4 x\,\sec^2 x = e^{\tan x}\sec^2 x\big(\tan^5 x + 5\tan^4 x\big).$

   Using $\sec^2 x = 1 + \tan^2 x$: …

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2. TG EAPCET 2026Set eng-2026-05-10-AN1 markMCQ
   Q.If $\int \frac{3e^x + 5e^{-x}}{1 - 4e^{-x}} dx = 3f(x) + \frac{5}{4}g(x) + \frac{53}{4}\log h(x) + c,\ f(0) = 1,\ g(0) = 0$ and $h(0) = 3$, then $f(1) + g(1) + h(1) =$ (A) $4$ (B) $-4$ (C) $3$ (D) $-3$

   ›Reveal solutionSolution

   The integral simplifies by rewriting the integrand in terms of $e^x$, splitting into partial fractions, and matching the given form to find $f(x)=x$, $g(x)=\log(1-4e^{-x})$, $h(x)=e^x-4$, then evaluating at $x=1$ gives $f(1)+g(1)+h(1)=4$, so the answer is (A).

   We are given:

   $\int \frac{3e^x + 5e^{-x}}{1 - 4e^{-x}} \, dx = 3f(x) + \frac{5}{4}g(x) + \frac{53}{4}\log h(x) + c,$

   with initial conditions $f(0)=1$, $g(0)=0$, $h(0)=3$. We need $f(1)+g(1)+h(1)$.

   Concept and intuition:

   The integrand mixes $e^x$ and $e^{-x}$. A natural first step is to rewrite everything in terms of $e^x$ (or $e^{-x}$) to simplify. Then the integral becomes a rational function in $e^x$, which we can integrate by splitting into simpler pieces. The given form of the antiderivative suggests that $f(x)$ will be something linear, $g(x)$ a logarithm, and $h(x)$ something inside a log. Matching terms will reveal each function explicitly.

   Step-by-step solution:

   1. Rewrite the integrand in terms of $e^x$. Multiply numerator and denominator by $e^x$ to clear the negative exponent:

   $\frac{3e^x + 5e^{-x}}{1 - 4e^{-x}} = \frac{3e^{2x} + 5}{e^x - 4}.$

   So the integral becomes

   $\int \frac{3e^{2x} + 5}{e^x - 4} \, dx.$

   2. Perform polynomial division (or split the fraction). Since the numerator is quadratic in $e^x$ and denominator linear, divide:

   $\frac{3e^{2x} + 5}{e^x - 4} = 3e^x + 12 + \frac{53}{e^x - 4}.$

   Check: $(e^x - 4)(3e^x + 12) = 3e^{2x} + 12e^x - 12e^x - 48 = 3e^{2x} - 48$, and we need $+5$, so remainder is $53$. Yes.

   3. Integrate term by term.

   $\int \left(3e^x + 12 + \frac{53}{e^x - 4}\right) dx = 3e^x + 12x + 53 \int \frac{dx}{e^x - 4}.$

   4. Handle the remaining integral. For $\int \frac{dx}{e^x - 4}$, substitute $u = e^x$, $du = e^x dx = u\,dx$, so $dx = \frac{du}{u}$:

   $\int \frac{dx}{e^x - 4} = \int \frac{du}{u(u-4)}.$

   Partial fractions:

   $\frac{1}{u(u-4)} = \frac{1}{4}\left(\frac{1}{u-4} - \frac{1}{u}\right).$

   Integrate:

   $\frac{1}{4} \left( \log|u-4| - \log|u| \right) = \frac{1}{4} \log\left|\frac{e^x - 4}{e^x}\right| = \frac{1}{4} \log|1 - 4e^{-x}|.$

   5. Assemble the full antiderivative.

   $\int \frac{3e^x + 5e^{-x}}{1 - 4e^{-x}} dx = 3e^x + 12x + 53 \cdot \frac{1}{4} \log|1 - 4e^{-x}| + c.$

   But the problem gives the form:

   $3f(x) + \frac{5}{4}g(x) + \frac{53}{4}\log h(x) + c.$

   Compare: The term $3e^x$ suggests $f(x) = e^x$? But then $f(0)=1$ matches. However, we also have $12x$ — that must be absorbed somewhere. Notice $12x = 3 \cdot 4x$, so we can write $3f(x) = 3e^x + 12x$ if $f(x) = e^x + 4x$. Check $f(0)=1+0=1$, good.

   Next, $\frac{5}{4}g(x)$ must match something; we have no term with coefficient $\frac{5}{4}$ yet. The only other term is $\frac{53}{4}\log|1-4e^{-x}|$. So likely $g(x)$ is something else and $h(x)$ is $|1-4e^{-x}|$? But then the coefficient $\frac{53}{4}$ matches, so $h(x) = |1-4e^{-x}|$. However $h(0)=|1-4|=3$, good.

   But then where is $\frac{5}{4}g(x)$? It seems missing — unless we split the log term differently. Actually, note that

   $\frac{53}{4}\log|1-4e^{-x}| = \frac{5}{4}\log|1-4e^{-x}| + \frac{48}{4}\log|1-4e^{-x}|.$

   The $\frac{48}{4}=12$ part could combine with $12x$? No, logs and polynomials are different. So perhaps the intended split is:

   $3f(x) = 3e^x + 12x \quad \Rightarrow \quad f(x)=e^x+4x,$

   and then

   $\frac{5}{4}g(x) + \frac{53}{4}\log h(x) = \frac{53}{4}\log|1-4e^{-x}|.$

   This forces $g(x)=0$ and $h(x)=|1-4e^{-x}|$, but then $g(0)=0$ holds, but the coefficient $\frac{5}{4}$ is unused. That seems inconsistent.

   Let’s re-examine: The given form has three terms: $3f(x)$, $\frac{5}{4}g(x)$, $\frac{53}{4}\log h(x)$. Our antiderivative has $3e^x$, $12x$, and $\frac{53}{4}\log|1-4e^{-x}|$. So we must match:

   $3f(x) = 3e^x + 12x \quad \Rightarrow \quad f(x)=e^x+4x,$

   $\frac{5}{4}g(x) = 0 \quad \Rightarrow \quad g(x)=0,$

   $\frac{53}{4}\log h(x) = \frac{53}{4}\log|1-4e^{-x}| \quad \Rightarrow \quad h(x)=|1-4e^{-x}|.$

   But then $g(x)=0$ identically, so $g(1)=0$. Then $f(1)=e+4$, $h(1)=|1-4e^{-1}| = |(e-4)/e| = (4-e)/e$? That is not a nice number, and sum won't be an integer. So this interpretation is wrong.

   The problem likely intends that the coefficients in front of $f$, $g$, and $\log h$ are fixed, and the functions themselves absorb the rest. In other words, we need to split the constant term differently. …

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3. TG EAPCET 2026Set eng-2026-05-10-FN1 markMCQ
   Q.If two roots of the equation $x^5 - 9x^4 + 27x^3 - 23x^2 - 24x + 36 = 0$ are repeated roots of multiplicity 2 and all the roots are integers then the sum of the squares of all different roots of the equation is (A) 14 (B) 29 (C) 26 (D) 21

   ›Reveal solutionSolution

   We use the Rational Root Theorem and synthetic division to find the integer roots of the quintic polynomial. Given that two roots have multiplicity 2 and all roots are integers, we systematically identify the distinct roots as $-1, 2,$ and $3$. The sum of the squares of these distinct roots is $\boxed{14}$.

   Concept and Intuition

   When dealing with polynomial equations, especially of higher degrees, finding roots can be challenging. However, this problem provides crucial information: all roots are integers, and there are specific multiplicities. This allows us to use powerful tools:

   1. Rational Root Theorem: If a polynomial with integer coefficients has a rational root $p/q$ (in simplest form), then $p$ must be a divisor of the constant term and $q$ must be a divisor of the leading coefficient. Since all roots are integers, $q$ must be $1$ (as the leading coefficient is $1$), meaning any integer root must be a divisor of the constant term. This significantly narrows down the possibilities for integer roots.
   2. Synthetic Division: Once a root $r$ is found, we know that $(x-r)$ is a factor of the polynomial. Synthetic division allows us to divide the polynomial by $(x-r)$ efficiently, yielding a depressed polynomial of one degree lower. This process can be repeated to find more roots.
   3. Multiplicity of Roots: A root $r$ has multiplicity $k$ if $(x-r)^k$ is a factor of the polynomial, but $(x-r)^{k+1}$ is not. This means $r$ is a root of the original polynomial, and also a root of the depressed polynomial obtained after dividing by $(x-r)$, and so on, $k$ times. In this problem, "two roots are repeated roots of multiplicity 2" means there are two distinct roots, say $a$ and $b$, such that $a$ appears twice and $b$ appears twice. The fifth root, $c$, must then appear once. So the roots are $a, a, b, b, c$.

   Our strategy will be to use the Rational Root Theorem to test integer divisors of the constant term. Once a root is found, we'll use synthetic division. If a root has multiplicity 2, it means we can divide by its corresponding factor twice.

   Step-by-step Solution

   1. Identify the polynomial and its properties:

      The given polynomial equation is $P(x) = x^5 - 9x^4 + 27x^3 - 23x^2 - 24x + 36 = 0$.

      It's a quintic polynomial, so it has 5 roots (counting multiplicity). We are told all roots are integers.

      We are also told that two roots are repeated roots of multiplicity 2. This means there are two distinct roots, say $r_1$ and $r_2$, each appearing twice. The fifth root, $r_3$, must be a simple root (multiplicity 1). So the set of roots is $\{r_1, r_1, r_2, r_2, r_3\}$.

   2. Apply the Rational Root Theorem:

      Since all roots are integers, they must be divisors of the constant term, $36$.

      The integer divisors of $36$ are $\pm 1, \pm 2, \pm 3, \pm 4, \pm 6, \pm 9, \pm 12, \pm 18, \pm 36$. We will test these values.

   3. Test for integer roots using synthetic division:

      Let's start by testing small integer values.

      • Test $x=1$: $P(1) = 1 - 9 + 27 - 23 - 24 + 36 = 8 \neq 0$. So $x=1$ is not a root.
      • Test $x=-1$: $P(-1) = (-1)^5 - 9(-1)^4 + 27(-1)^3 - 23(-1)^2 - 24(-1) + 36$ $P(-1) = -1 - 9 - 27 - 23 + 24 + 36 = 0$. So, $x=-1$ is a root.

   4. Perform synthetic division with $x=-1$:

      Dividing $P(x)$ by $(x+1)$:

      -1 | 1  -9   27  -23  -24   36
         |    -1   10  -37   60  -36
         ---------------------------
           1 -10   37  -60   36    0
      

      The depressed polynomial is $Q_1(x) = x^4 - 10x^3 + 37x^2 - 60x + 36$.

   5. Check multiplicity of $x=-1$:

      We need to check if $x=-1$ is a root of $Q_1(x)$.

      $Q_1(-1) = (-1)^4 - 10(-1)^3 + 37(-1)^2 - 60(-1) + 36$

      $Q_1(-1) = 1 + 10 + 37 + 60 + 36 = 144 \neq 0$.

      Since $Q_1(-1) \neq 0$, $x=-1$ is not a repeated root. This means $x=-1$ is a simple root (multiplicity 1), so it must be $r_3$.

      The remaining four roots, $r_1, r_1, r_2, r_2$, must be the roots of $Q_1(x) = x^4 - 10x^3 + 37x^2 - 60x + 36 = 0$.

   6. Find roots of $Q_1(x)$: …

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4. TG EAPCET 2025Set eng-2025-05-02-FN1 markMCQ
   Q.$\int \left( \frac{1 - \log x}{1 + (\log x)^2} \right)^2 dx =$ (A) $\frac{1}{1 + (\log x)^2} + c$ (B) $\frac{\log x}{1 + (\log x)^2} + c$ (C) $\frac{x}{1 + (\log x)^2} + c$ (D) $\frac{x^2}{1 + (\log x)^2} + c$

   ›Reveal solutionSolution

   The key is to notice that the integrand is the derivative of a rational function of $\log x$ after a clever substitution. The integral simplifies to $\frac{x}{1+(\log x)^2}+C$, so the correct option is (C).

   We start with the integral

   $I = \int \left( \frac{1 - \log x}{1 + (\log x)^2} \right)^2 dx.$

   The presence of $\log x$ suggests the substitution $t = \log x$, so $x = e^t$ and $dx = e^t dt$. Then the integral becomes

   $I = \int \left( \frac{1 - t}{1 + t^2} \right)^2 e^t \, dt.$

   Now we have a product of a rational function in $t$ and $e^t$. This often hints at integration by parts or recognizing a derivative of the form $\frac{d}{dt}\left( \frac{e^t}{1+t^2} \right)$. Let’s check:

   $\frac{d}{dt}\left( \frac{e^t}{1+t^2} \right) = \frac{e^t(1+t^2) - e^t(2t)}{(1+t^2)^2} = \frac{e^t(1 - 2t + t^2)}{(1+t^2)^2} = \frac{e^t(1-t)^2}{(1+t^2)^2}.$

   That is exactly $e^t \left( \frac{1-t}{1+t^2} \right)^2$, which is our integrand! So

   $I = \int \frac{d}{dt}\left( \frac{e^t}{1+t^2} \right) dt = \frac{e^t}{1+t^2} + C.$

   Substituting back $t = \log x$, we get $e^t = x$ and $t^2 = (\log x)^2$, so

   $I = \frac{x}{1 + (\log x)^2} + C.$ …

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5. TG EAPCET 2025Set eng-2025-05-02-FN1 markMCQ
   Q.If $\int (x+2)\sqrt{x^2-x+2}\,dx = \frac{1}{3}f(x) + \frac{5}{8}g(x) + \frac{35}{16}h(x) + c$ then $f(-1) + g(-1) + h\left(\frac{1}{2}\right) =$ (A) $-4$ (B) $2$ (C) $4$ (D) $-2$

   ›Reveal solutionSolution

   Complete the square and split the integral; matching the given form gives $f(x)=(x^2-x+2)^{3/2}$, $g(x)=(2x-1)\sqrt{x^2-x+2}$, $h(x)=\sinh^{-1}\!\frac{2x-1}{\sqrt7}$, and $f(-1)+g(-1)+h(\tfrac12)=2$ — option (B).

   Complete the square. $x^2-x+2=\left(x-\tfrac12\right)^2+\tfrac74$. Let $u=x-\tfrac12$, so $x+2=u+\tfrac52$:

   $\int\left(u+\tfrac52\right)\sqrt{u^2+\tfrac74}\,du.$

   First part. With $t=u^2+\tfrac74$,

   $\int u\sqrt{u^2+\tfrac74}\,du=\tfrac13\left(u^2+\tfrac74\right)^{3/2}=\tfrac13\left(x^2-x+2\right)^{3/2}.$

   Second part. Using $\int\sqrt{u^2+a^2}\,du=\dfrac{u}{2}\sqrt{u^2+a^2}+\dfrac{a^2}{2}\sinh^{-1}\!\dfrac{u}{a}$ with $a^2=\tfrac74$:

   $\tfrac52\int\sqrt{u^2+\tfrac74}\,du=\frac{5}{4}\left(x-\tfrac12\right)\sqrt{x^2-x+2}+\frac{35}{16}\sinh^{-1}\!\frac{2x-1}{\sqrt7}.$ …

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6. TG EAPCET 2025Set eng-2025-05-03-AN1 markMCQ
   Q.$\int \frac{1}{(2\cos x + \sin x)^2} dx =$ (A) $\frac{1}{2+\tan x} + c$ (B) $-\frac{1}{2\tan x + 1} + c$ (C) $\frac{\cos x}{\cos x + 2\sin x} + c$ (D) $-\frac{\cos x}{2\cos x + \sin x} + c$

   ›Reveal solutionSolution

   The integral simplifies by dividing numerator and denominator by $\cos^2 x$, turning it into a standard form $\int \frac{du}{(2+u)^2}$ after substitution $u = \tan x$. The result matches option (D).

   We want to evaluate

   $\int \frac{1}{(2\cos x + \sin x)^2} \, dx.$

   The denominator is a linear combination of $\cos x$ and $\sin x$, squared. A natural instinct is to try a substitution that simplifies such expressions: dividing by $\cos^2 x$ often works because it turns the denominator into something like $(a + b\tan x)^2$ times $\sec^2 x$, and $\sec^2 x$ is the derivative of $\tan x$.

   Why this works:

   If we rewrite the integrand as

   $\frac{1}{(2\cos x + \sin x)^2} = \frac{1}{\cos^2 x \, (2 + \tan x)^2},$

   then we have $\frac{1}{\cos^2 x} = \sec^2 x$, which is exactly $d(\tan x)/dx$. So the integral becomes $\int \frac{\sec^2 x}{(2+\tan x)^2} dx$, and substituting $u = \tan x$ gives a simple power rule integral.

   Let’s do it step by step.

   1. Factor $\cos^2 x$ from the denominator

   $(2\cos x + \sin x)^2 = \cos^2 x \, (2 + \tan x)^2.$

   This is valid because $\sin x / \cos x = \tan x$, provided $\cos x \neq 0$ (we can handle singularities separately; the antiderivative will be valid on intervals where $\cos x \neq 0$).

   2. Rewrite the integrand

   $\frac{1}{(2\cos x + \sin x)^2} = \frac{1}{\cos^2 x \, (2 + \tan x)^2} = \frac{\sec^2 x}{(2 + \tan x)^2}.$

   3. Substitute $u = \tan x$ Then $du = \sec^2 x \, dx$, so the integral becomes

   $\int \frac{du}{(2 + u)^2}.$

   4. Integrate

   $\int \frac{du}{(2+u)^2} = -\frac{1}{2+u} + C.$

   5. Back-substitute $u = \tan x$

   $-\frac{1}{2 + \tan x} + C.$

   6. Rewrite in terms of $\cos x$ and $\sin x$

   $-\frac{1}{2 + \frac{\sin x}{\cos x}} = -\frac{1}{\frac{2\cos x + \sin x}{\cos x}} = -\frac{\cos x}{2\cos x + \sin x}.$

   So the antiderivative is

   $-\frac{\cos x}{2\cos x + \sin x} + C.$

   Tip

   A common pitfall is forgetting the minus sign from $\int u^{-2} du = -u^{-1}$. Also, some might try to use the tangent half-angle substitution, but that’s overkill here — dividing by $\cos^2 x$ is much faster. …

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7. TG EAPCET 2024Set eng-2024-05-10-AN1 markMCQ
   Q.If $\int \frac{1}{x^4 + 8x^2 + 9} dx = \frac{1}{k} \left[ \frac{1}{\sqrt{14}} \tan^{-1}(f(x)) - \frac{1}{\sqrt{2}} \tan^{-1}(g(x)) \right] + c$, then $\sqrt{\frac{k}{2} + f(\sqrt{3}) + g(1)} =$ (A) $3 - 2\sqrt{2}$ (B) $\sqrt{2} - 1$ (C) $\sqrt{3} + 2\sqrt{2}$ (D) $\sqrt{2} + 1$

   ›Reveal solutionSolution

   The key is to factor the quartic denominator into two quadratics, decompose the integrand via partial fractions, integrate each term to match the given form, then identify $k$, $f(x)$, and $g(x)$ and evaluate the expression at the given points.

   We are given

   $\int \frac{1}{x^4 + 8x^2 + 9} \, dx = \frac{1}{k} \left[ \frac{1}{\sqrt{14}} \tan^{-1}(f(x)) - \frac{1}{\sqrt{2}} \tan^{-1}(g(x)) \right] + c$

   and need to find $\sqrt{\frac{k}{2} + f(\sqrt{3}) + g(1)}$.

   ---

   1. Factor the denominator

   The denominator $x^4 + 8x^2 + 9$ is a quadratic in $x^2$. Let $t = x^2$:

   $t^2 + 8t + 9 = 0 \quad\Rightarrow\quad t = -4 \pm \sqrt{7}.$

   So

   $x^4 + 8x^2 + 9 = (x^2 + 4 - \sqrt{7})(x^2 + 4 + \sqrt{7}).$

   We can also write it as

   $(x^2 + a)(x^2 + b) \quad\text{with}\quad a+b=8,\; ab=9.$

   Solving gives $a,b = 4 \pm \sqrt{7}$. So

   $x^4 + 8x^2 + 9 = (x^2 + 4 - \sqrt{7})(x^2 + 4 + \sqrt{7}).$

   ---

   2. Partial fraction decomposition

   We want

   $\frac{1}{(x^2 + A)(x^2 + B)} = \frac{Px + Q}{x^2 + A} + \frac{Rx + S}{x^2 + B}.$

   Since the numerator is constant and denominators are even, the numerators will be constants (no $x$ terms). So

   $\frac{1}{(x^2 + A)(x^2 + B)} = \frac{C}{x^2 + A} + \frac{D}{x^2 + B}.$

   Multiply through:

   $1 = C(x^2 + B) + D(x^2 + A) = (C + D)x^2 + (CB + DA).$

   Thus

   $C + D = 0,\quad CB + DA = 1.$

   From $D = -C$, the second gives $C(B - A) = 1$, so

   $C = \frac{1}{B - A},\quad D = \frac{1}{A - B}.$

   Here $A = 4 - \sqrt{7}$, $B = 4 + \sqrt{7}$, so $B - A = 2\sqrt{7}$. Hence

   $C = \frac{1}{2\sqrt{7}},\quad D = -\frac{1}{2\sqrt{7}}.$

   Thus

   $\frac{1}{x^4 + 8x^2 + 9} = \frac{1}{2\sqrt{7}} \left( \frac{1}{x^2 + 4 - \sqrt{7}} - \frac{1}{x^2 + 4 + \sqrt{7}} \right).$

   ---

   3. Integrate each term

   Recall

   $\int \frac{1}{x^2 + p^2} \, dx = \frac{1}{p} \tan^{-1}\left(\frac{x}{p}\right) + c.$

   For the first term: $p^2 = 4 - \sqrt{7}$. But note $4 - \sqrt{7} > 0$?

   $\sqrt{7} \approx 2.6458$, so $4 - \sqrt{7} \approx 1.3542 > 0$. So $p_1 = \sqrt{4 - \sqrt{7}}$.

   For the second: $p_2 = \sqrt{4 + \sqrt{7}}$.

   Thus

   $\int \frac{1}{x^2 + 4 - \sqrt{7}} \, dx = \frac{1}{\sqrt{4 - \sqrt{7}}} \tan^{-1}\left( \frac{x}{\sqrt{4 - \sqrt{7}}} \right)$

   $\int \frac{1}{x^2 + 4 + \sqrt{7}} \, dx = \frac{1}{\sqrt{4 + \sqrt{7}}} \tan^{-1}\left( \frac{x}{\sqrt{4 + \sqrt{7}}} \right).$

   So

   $\int \frac{dx}{x^4 + 8x^2 + 9} = \frac{1}{2\sqrt{7}} \left[ \frac{1}{\sqrt{4 - \sqrt{7}}} \tan^{-1}\left( \frac{x}{\sqrt{4 - \sqrt{7}}} \right) - \frac{1}{\sqrt{4 + \sqrt{7}}} \tan^{-1}\left( \frac{x}{\sqrt{4 + \sqrt{7}}} \right) \right] + c.$

   ---

   4. Match to given form

   The given form is

   $\frac{1}{k} \left[ \frac{1}{\sqrt{14}} \tan^{-1}(f(x)) - \frac{1}{\sqrt{2}} \tan^{-1}(g(x)) \right].$

   Compare coefficients:

   $\frac{1}{2\sqrt{7}} \cdot \frac{1}{\sqrt{4 - \sqrt{7}}} \quad\text{should equal}\quad \frac{1}{k} \cdot \frac{1}{\sqrt{14}}.$

   Also

   $\frac{1}{2\sqrt{7}} \cdot \frac{1}{\sqrt{4 + \sqrt{7}}} \quad\text{should equal}\quad \frac{1}{k} \cdot \frac{1}{\sqrt{2}}.$

   And

   $f(x) = \frac{x}{\sqrt{4 - \sqrt{7}}},\quad g(x) = \frac{x}{\sqrt{4 + \sqrt{7}}}.$

   ---

   5. Find $k$

   From the second coefficient match:

   $\frac{1}{2\sqrt{7} \sqrt{4 + \sqrt{7}}} = \frac{1}{k \sqrt{2}}.$

   So

   $k = \frac{2\sqrt{7} \sqrt{4 + \sqrt{7}}}{\sqrt{2}} = \sqrt{14} \sqrt{4 + \sqrt{7}}.$

   Check with the first:

   $\frac{1}{2\sqrt{7} \sqrt{4 - \sqrt{7}}} = \frac{1}{k \sqrt{14}} \quad\Rightarrow\quad k = \frac{2\sqrt{7} \sqrt{4 - \sqrt{7}} \sqrt{14}}{1} = 2\sqrt{7}\sqrt{4 - \sqrt{7}}\sqrt{14}.$

   Simplify: $\sqrt{14} = \sqrt{2}\sqrt{7}$, so $2\sqrt{7}\sqrt{7}\sqrt{2}\sqrt{4 - \sqrt{7}} = 14\sqrt{2}\sqrt{4 - \sqrt{7}}$.

   But from the other we got $k = \sqrt{14}\sqrt{4 + \sqrt{7}}$. For consistency, these must be equal — and indeed they are because $(4 - \sqrt{7})(4 + \sqrt{7}) = 9$, so

   $\sqrt{4 - \sqrt{7}} = \frac{\sqrt{9}}{\sqrt{4 + \sqrt{7}}} = \frac{3}{\sqrt{4 + \sqrt{7}}}.$

   Then $14\sqrt{2} \cdot \frac{3}{\sqrt{4 + \sqrt{7}}} = \frac{42\sqrt{2}}{\sqrt{4 + \sqrt{7}}}$. That doesn’t match $\sqrt{14}\sqrt{4 + \sqrt{7}}$ unless we check numeric:

   $\sqrt{4 + \sqrt{7}} \approx \sqrt{6.6458} \approx 2.578$, $\sqrt{14} \approx 3.742$, product $\approx 9.65$.

   The other: $42\sqrt{2} \approx 59.4$, divided by 2.578 $\approx 23.05$. So they differ — meaning I made an algebra slip. Let’s re-derive carefully.

   ---

   6. Correct matching

   We have

   $\frac{1}{2\sqrt{7}} \cdot \frac{1}{\sqrt{4 - \sqrt{7}}} = \frac{1}{k \sqrt{14}} \quad\Rightarrow\quad k = \frac{2\sqrt{7} \sqrt{4 - \sqrt{7}} \sqrt{14}}{1}.$

   Simplify $\sqrt{14} = \sqrt{2}\sqrt{7}$, so

   $k = 2\sqrt{7} \cdot \sqrt{4 - \sqrt{7}} \cdot \sqrt{2}\sqrt{7} = 2 \cdot 7 \cdot \sqrt{2} \cdot \sqrt{4 - \sqrt{7}} = 14\sqrt{2} \sqrt{4 - \sqrt{7}}.$

   From the second:

   $\frac{1}{2\sqrt{7} \sqrt{4 + \sqrt{7}}} = \frac{1}{k \sqrt{2}} \quad\Rightarrow\quad k = 2\sqrt{7} \sqrt{4 + \sqrt{7}} \sqrt{2} = 2\sqrt{14} \sqrt{4 + \sqrt{7}}.$

   These two expressions for $k$ must be equal:

   $14\sqrt{2} \sqrt{4 - \sqrt{7}} = 2\sqrt{14} \sqrt{4 + \sqrt{7}}.$

   Divide both sides by $2\sqrt{2}$: left becomes $7 \sqrt{4 - \sqrt{7}}$, right becomes $\sqrt{7} \sqrt{4 + \sqrt{7}}$. Square: left $49(4 - \sqrt{7}) = 196 - 49\sqrt{7}$, right $7(4 + \sqrt{7}) = 28 + 7\sqrt{7}$. Not equal — so the given form’s coefficients $\frac{1}{\sqrt{14}}$ and $\frac{1}{\sqrt{2}}$ are not directly the arctan coefficients; they are inside the bracket multiplied by $1/k$. So we must solve for $k$ from either equation; they will give the same $k$ if the given form is correct. Let’s pick the second:

   $k = 2\sqrt{14} \sqrt{4 + \sqrt{7}}.$

   We can simplify $\sqrt{4 + \sqrt{7}}$. Note $( \sqrt{4 + \sqrt{7}} )^2 = 4 + \sqrt{7}$. Sometimes it’s expressible as $\sqrt{a} + \sqrt{b}$. Try:

   Let $\sqrt{4 + \sqrt{7}} = \sqrt{m} + \sqrt{n}$. Squaring: $m + n + 2\sqrt{mn} = 4 + \sqrt{7}$. So $m+n=4$, $2\sqrt{mn} = \sqrt{7} \Rightarrow 4mn = 7 \Rightarrow mn = 7/4$. Solve: $m,n$ are roots of $t^2 - 4t + 7/4 = 0$ → $t = \frac{4 \pm \sqrt{16 - 7}}{2} = \frac{4 \pm 3}{2}$. So $m = 7/2, n = 1/2$. Thus

   $\sqrt{4 + \sqrt{7}} = \sqrt{\frac{7}{2}} + \sqrt{\frac{1}{2}} = \frac{\sqrt{14} + \sqrt{2}}{2}.$

   Similarly, $\sqrt{4 - \sqrt{7}} = \frac{\sqrt{14} - \sqrt{2}}{2}$.

   So

   $k = 2\sqrt{14} \cdot \frac{\sqrt{14} + \sqrt{2}}{2} = \sqrt{14}(\sqrt{14} + \sqrt{2}) = 14 + \sqrt{28} = 14 + 2\sqrt{7}.$

   Check with the other expression: $k = 14\sqrt{2} \cdot \frac{\sqrt{14} - \sqrt{2}}{2} = 7\sqrt{2}(\sqrt{14} - \sqrt{2}) = 7\sqrt{28} - 7\cdot 2 = 7\cdot 2\sqrt{7} - 14 = 14\sqrt{7} - 14$. That’s different — so the given form’s constants $\sqrt{14}$ and $\sqrt{2}$ are swapped relative to our derivation? Let’s re-examine: In the given form, the first arctan has coefficient $1/\sqrt{14}$, the second has $1/\sqrt{2}$. In our integral, the first term (with $4 - \sqrt{7}$) had coefficient $1/(2\sqrt{7}\sqrt{4 - \sqrt{7}})$. Compute that:

   $\frac{1}{2\sqrt{7}} \cdot \frac{1}{\sqrt{4 - \sqrt{7}}} = \frac{1}{2\sqrt{7}} \cdot \frac{2}{\sqrt{14} - \sqrt{2}} = \frac{1}{\sqrt{7}(\sqrt{14} - \sqrt{2})}.$

   Rationalize: multiply numerator and denominator by $\sqrt{14} + \sqrt{2}$:

   $= \frac{\sqrt{14} + \sqrt{2}}{\sqrt{7}(14 - 2)} = \frac{\sqrt{14} + \sqrt{2}}{12\sqrt{7}} = \frac{\sqrt{2}(\sqrt{7} + 1)}{12\sqrt{7}}.$

   That’s not obviously $1/(k\sqrt{14})$. Instead, match directly:

   We want

   $\frac{1}{2\sqrt{7}\sqrt{4 - \sqrt{7}}} = \frac{1}{k\sqrt{14}} \quad\Rightarrow\quad k = \frac{2\sqrt{7}\sqrt{4 - \sqrt{7}}\sqrt{14}}{1} = 2\sqrt{98}\sqrt{4 - \sqrt{7}} = 14\sqrt{2}\sqrt{4 - \sqrt{7}}.$

   Using $\sqrt{4 - \sqrt{7}} = (\sqrt{14} - \sqrt{2})/2$, we get

   $k = 14\sqrt{2} \cdot \frac{\sqrt{14} - \sqrt{2}}{2} = 7\sqrt{2}(\sqrt{14} - \sqrt{2}) = 7\sqrt{28} - 14 = 14\sqrt{7} - 14.$

   From the second term:

   $\frac{1}{2\sqrt{7}\sqrt{4 + \sqrt{7}}} = \frac{1}{k\sqrt{2}} \quad\Rightarrow\quad k = 2\sqrt{7}\sqrt{4 + \sqrt{7}}\sqrt{2} = 2\sqrt{14}\sqrt{4 + \sqrt{7}} = 2\sqrt{14}\cdot \frac{\sqrt{14} + \sqrt{2}}{2} = 14 + 2\sqrt{7}.$

   These are not equal — meaning the given form’s coefficients $\frac{1}{\sqrt{14}}$ and $\frac{1}{\sqrt{2}}$ are swapped relative to our natural ordering. So we must assign:

   The term with $\frac{1}{\sqrt{14}}$ corresponds to the larger denominator (since $\sqrt{4 + \sqrt{7}} > \sqrt{4 - \sqrt{7}}$ and $1/\sqrt{14} < 1/\sqrt{2}$). So let

   $\frac{1}{k\sqrt{14}} = \frac{1}{2\sqrt{7}\sqrt{4 + \sqrt{7}}}, \quad \frac{1}{k\sqrt{2}} = \frac{1}{2\sqrt{7}\sqrt{4 - \sqrt{7}}}.$

   Then from the first:

   $k = \frac{2\sqrt{7}\sqrt{4 + \sqrt{7}}\sqrt{14}}{1} = 2\sqrt{98}\sqrt{4 + \sqrt{7}} = 14\sqrt{2}\sqrt{4 + \sqrt{7}}.$

   Using $\sqrt{4 + \sqrt{7}} = (\sqrt{14} + \sqrt{2})/2$,

   $k = 14\sqrt{2} \cdot \frac{\sqrt{14} + \sqrt{2}}{2} = 7\sqrt{2}(\sqrt{14} + \sqrt{2}) = 7\sqrt{28} + 14 = 14\sqrt{7} + 14.$

   From the second:

   $k = \frac{2\sqrt{7}\sqrt{4 - \sqrt{7}}\sqrt{2}}{1} = 2\sqrt{14}\sqrt{4 - \sqrt{7}} = 2\sqrt{14} \cdot \frac{\sqrt{14} - \sqrt{2}}{2} = 14 - 2\sqrt{7}.$

   Still not equal — so the given form’s constants are not directly the arctan coefficients; they are scaled by $1/k$ and the arctan arguments absorb the rest. Actually, the given form has $f(x)$ and $g(x)$ as the arguments, not necessarily $x/p$. So we can set

   $f(x) = \frac{x}{\sqrt{4 + \sqrt{7}}}, \quad g(x) = \frac{x}{\sqrt{4 - \sqrt{7}}}$

   and then match coefficients accordingly. Let’s do that systematically.

   ---

   7. Systematic matching

   We have

   $\int = \frac{1}{2\sqrt{7}} \left[ \frac{1}{\sqrt{4 - \sqrt{7}}} \tan^{-1}\left( \frac{x}{\sqrt{4 - \sqrt{7}}} \right) - \frac{1}{\sqrt{4 + \sqrt{7}}} \tan^{-1}\left( \frac{x}{\sqrt{4 + \sqrt{7}}} \right) \right].$

   We want this equal to

   $\frac{1}{k} \left[ \frac{1}{\sqrt{14}} \tan^{-1}(f(x)) - \frac{1}{\sqrt{2}} \tan^{-1}(g(x)) \right].$

   So we can identify:

   $\frac{1}{k\sqrt{14}} = \frac{1}{2\sqrt{7}\sqrt{4 - \sqrt{7}}}, \quad f(x) = \frac{x}{\sqrt{4 - \sqrt{7}}}$

   $\frac{1}{k\sqrt{2}} = \frac{1}{2\sqrt{7}\sqrt{4 + \sqrt{7}}}, \quad g(x) = \frac{x}{\sqrt{4 + \sqrt{7}}}.$

   Now solve for $k$ from the first:

   $k = \frac{2\sqrt{7}\sqrt{4 - \sqrt{7}}\sqrt{14}}{1} = 2\sqrt{98}\sqrt{4 - \sqrt{7}} = 14\sqrt{2}\sqrt{4 - \sqrt{7}}.$

   Using $\sqrt{4 - \sqrt{7}} = (\sqrt{14} - \sqrt{2})/2$,

   $k = 14\sqrt{2} \cdot \frac{\sqrt{14} - \sqrt{2}}{2} = 7\sqrt{2}(\sqrt{14} - \sqrt{2}) = 7\sqrt{28} - 14 = 14\sqrt{7} - 14.$

   From the second:

   $k = \frac{2\sqrt{7}\sqrt{4 + \sqrt{7}}\sqrt{2}}{1} = 2\sqrt{14}\sqrt{4 + \sqrt{7}} = 2\sqrt{14} \cdot \frac{\sqrt{14} + \sqrt{2}}{2} = 14 + 2\sqrt{7}.$

   These are different — so the given form’s constants $\sqrt{14}$ and $\sqrt{2}$ are swapped relative to this assignment. So swap: let

   $\frac{1}{k\sqrt{2}} = \frac{1}{2\sqrt{7}\sqrt{4 - \sqrt{7}}}, \quad \frac{1}{k\sqrt{14}} = \frac{1}{2\sqrt{7}\sqrt{4 + \sqrt{7}}}.$

   Then

   $k = \frac{2\sqrt{7}\sqrt{4 - \sqrt{7}}\sqrt{2}}{1} = 2\sqrt{14}\sqrt{4 - \sqrt{7}} = 2\sqrt{14} \cdot \frac{\sqrt{14} - \sqrt{2}}{2} = 14 - 2\sqrt{7}.$

   And

   $$$$

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8. TG EAPCET 2024Set eng-2024-05-11-FN1 markMCQ
   Q.

   [!FORMULA] $\int \frac{dx}{(x+1)\sqrt{x^2+4}} =$

   (A) $\frac{1}{2} \sqrt{\frac{x+1}{x+2}} + c$ (B) $\log \left| \frac{x+2}{x+1} \right| + c$ (C) $-\frac{1}{\sqrt{5}} \sinh^{-1} \left( \frac{4-x}{2(x+1)} \right) + c$ (D) $-\frac{1}{\sqrt{5}} \cosh^{-1} \left( \frac{4+x}{2(x-1)} \right) + c$

   ›Reveal solutionSolution

   The integral $\int \frac{dx}{(x+1)\sqrt{x^2+4}}$ is solved by the substitution $x+1 = \frac{1}{t}$, which transforms it into a standard form, leading to the result $-\frac{1}{\sqrt{5}} \sinh^{-1}\left( \frac{4-x}{2(x+1)} \right) + c$, so the correct option is (C).

   The key insight: when you have a linear factor $(x+1)$ multiplied by a square root of a quadratic, a substitution like $x+1 = \frac{1}{t}$ often rationalizes the expression. This works because it turns the denominator into something like $t^2$ times a simpler radical, and the quadratic $x^2+4$ becomes a function of $t$ that fits a standard inverse hyperbolic form.

   Let’s work through it step by step.

   1. Set up the substitution Let $x+1 = \frac{1}{t}$, so $x = \frac{1}{t} - 1$. Then $dx = -\frac{1}{t^2} \, dt$. The integral becomes:

   $\int \frac{dx}{(x+1)\sqrt{x^2+4}} = \int \frac{-\frac{1}{t^2} \, dt}{\frac{1}{t} \cdot \sqrt{\left(\frac{1}{t} - 1\right)^2 + 4}}.$

   2. Simplify the denominator The factor $\frac{1}{t}$ in the denominator cancels one power of $t$ from $dx$:

   $\frac{-\frac{1}{t^2}}{\frac{1}{t}} = -\frac{1}{t}.$

   So the integral is:

   $\int -\frac{1}{t} \cdot \frac{dt}{\sqrt{\left(\frac{1}{t} - 1\right)^2 + 4}}.$

   3. Simplify the expression inside the square root Compute:

   $\left(\frac{1}{t} - 1\right)^2 + 4 = \frac{1}{t^2} - \frac{2}{t} + 1 + 4 = \frac{1}{t^2} - \frac{2}{t} + 5.$

   Multiply numerator and denominator inside the square root by $t^2$ to clear fractions:

   $\sqrt{\frac{1}{t^2} - \frac{2}{t} + 5} = \frac{\sqrt{1 - 2t + 5t^2}}{|t|}.$

   Since we are dealing with an indefinite integral, we can assume $t>0$ (or handle sign with absolute values later; the result will involve a logarithm/hyperbolic function). So $|t| = t$.

   4. Rewrite the integral Substituting back:

   $\int -\frac{1}{t} \cdot \frac{dt}{\frac{\sqrt{1 - 2t + 5t^2}}{t}} = \int -\frac{1}{t} \cdot \frac{t}{\sqrt{1 - 2t + 5t^2}} \, dt = \int -\frac{dt}{\sqrt{1 - 2t + 5t^2}}.$

   5. Complete the square in the quadratic $1 - 2t + 5t^2 = 5\left(t^2 - \frac{2}{5}t\right) + 1 = 5\left[\left(t - \frac{1}{5}\right)^2 - \frac{1}{25}\right] + 1 = 5\left(t - \frac{1}{5}\right)^2 - \frac{1}{5} + 1 = 5\left(t - \frac{1}{5}\right)^2 + \frac{4}{5}$. Factor out the 5:

   $\sqrt{1 - 2t + 5t^2} = \sqrt{5} \sqrt{\left(t - \frac{1}{5}\right)^2 + \frac{4}{25}}.$

   6. Standard hyperbolic sine inverse form Recall: $\int \frac{du}{\sqrt{u^2 + a^2}} = \sinh^{-1}\left(\frac{u}{a}\right) + c$. Here $u = t - \frac{1}{5}$, $a = \frac{2}{5}$. So: $$$$

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9. TG EAPCET 2023Set eng-2023-05-13-FN1 markMCQ
   Q.$\int \frac{\sec^2 x}{(\sec x + \tan x)^2} dx =$ (A) $\frac{3 + (\sec x + \tan x)^2}{2(\sec x + \tan x)^3} + C$ (B) $-\frac{1 + 3(\sec x + \tan x)^2}{6(\sec x + \tan x)^3} + C$ (C) $-\frac{3 + (\sec x + \tan x)^2}{2(\sec x + \tan x)^3} + C$ (D) $-\frac{1 + (\sec x + \tan x)}{3(\sec x + \tan x)^2} + C$

   ›Reveal solutionSolution

   The integral simplifies by substituting $u = \sec x + \tan x$, which turns the integrand into a simple power of $u$. The result is $-\frac{1 + 3(\sec x + \tan x)^2}{6(\sec x + \tan x)^3} + C$, matching option (B).

   Concept & Intuition

   When you see $\sec x$ and $\tan x$ together, especially squared in the denominator, a classic trick is to recall that

   $\frac{d}{dx}(\sec x + \tan x) = \sec x \tan x + \sec^2 x = \sec x (\sec x + \tan x).$

   This suggests that $\sec x + \tan x$ is a natural substitution. The numerator $\sec^2 x$ is almost the derivative of something related — we just need to express everything in terms of $u = \sec x + \tan x$. This substitution elegantly collapses the messy trigonometric integral into a rational function of $u$.

   Step-by-step solution

   1. Set up the substitution Let

   $u = \sec x + \tan x.$

   Then differentiate:

   $du = (\sec x \tan x + \sec^2 x)\, dx = \sec x (\sec x + \tan x)\, dx = \sec x \cdot u \, dx.$

   So

   $dx = \frac{du}{\sec x \cdot u}.$

   2. Express $\sec x$ in terms of $u$ We need $\sec x$ alone. A useful identity:

   $\sec x - \tan x = \frac{1}{\sec x + \tan x} = \frac{1}{u}.$

   Adding and subtracting:

   $\sec x = \frac{1}{2}\left( u + \frac{1}{u} \right), \quad \tan x = \frac{1}{2}\left( u - \frac{1}{u} \right).$

   So

   $\sec x = \frac{u^2 + 1}{2u}.$

   3. Rewrite the integral The integral is

   $I = \int \frac{\sec^2 x}{(\sec x + \tan x)^2} \, dx = \int \frac{\sec^2 x}{u^2} \, dx.$

   Substitute $dx = \frac{du}{\sec x \cdot u}$:

   $I = \int \frac{\sec^2 x}{u^2} \cdot \frac{du}{\sec x \cdot u} = \int \frac{\sec x}{u^3} \, du.$

   Now replace $\sec x$ with $\frac{u^2 + 1}{2u}$:

   $I = \int \frac{u^2 + 1}{2u} \cdot \frac{1}{u^3} \, du = \frac12 \int \frac{u^2 + 1}{u^4} \, du.$

   4. Integrate term by term

   $\frac{u^2 + 1}{u^4} = u^{-2} + u^{-4}.$

   So

   $$$$

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10. TG EAPCET 2023Set eng-2023-05-13-AN1 markMCQ
    Q.

    [!FORMULA] $\int \frac{1}{\left(x + \frac{2}{x}\right)\sqrt{x^4 + 4x^2 + 3}} \, dx =$

    (A) $\frac{1}{2} \sec^{-1}(x^2 + 2) + c$ (B) $-\cosech^{-1}(x^2 + 2) + c$ (C) $\frac{1}{2} \tan^{-1}\left(x + \frac{2}{x}\right) + c$ (D) $-\frac{1}{2} \cot^{-1}\left(x + \frac{2}{x}\right) + c$

    ›Reveal solutionSolution

    The key is to rewrite the integrand by factoring $x^2$ from the square root and substituting $t = x^2$, leading to a standard inverse secant form. The correct answer is (A).

    ---

    The problem looks messy at first: a compound denominator $x + \frac{2}{x}$ and a quartic under the square root. But the structure hints at a substitution that simplifies the square root into something like $\sqrt{(x^2+2)^2 - 1}$. That’s exactly the form that suggests an inverse secant (or inverse hyperbolic secant) result. The trick is to notice that $x^4 + 4x^2 + 3 = (x^2+2)^2 - 1$, and that the factor $x + \frac{2}{x}$ can be expressed in terms of $x^2$.

    ---

    1. Simplify the square root

    $x^4 + 4x^2 + 3 = (x^4 + 4x^2 + 4) - 1 = (x^2 + 2)^2 - 1.$

    So the integrand becomes

    $\frac{1}{\left(x + \frac{2}{x}\right) \sqrt{(x^2+2)^2 - 1}}.$

    2. Rewrite the denominator factor

    $x + \frac{2}{x} = \frac{x^2 + 2}{x}.$

    Hence the integral is

    $\int \frac{1}{\frac{x^2+2}{x} \cdot \sqrt{(x^2+2)^2 - 1}} \, dx = \int \frac{x}{(x^2+2) \sqrt{(x^2+2)^2 - 1}} \, dx.$

    3. Substitute $t = x^2$ Then $dt = 2x \, dx$, so $x \, dx = \frac{dt}{2}$. Also $x^2+2 = t+2$. The integral becomes

    $\int \frac{1}{(t+2) \sqrt{(t+2)^2 - 1}} \cdot \frac{dt}{2} = \frac12 \int \frac{dt}{(t+2) \sqrt{(t+2)^2 - 1}}.$

    4. Recognize the standard form Recall that $\int \frac{du}{u \sqrt{u^2 - 1}} = \sec^{-1} |u| + C.$ …

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11. TG EAPCET 2022Set eng-2022-07-18-AN1 markMCQ
    Q.If $\int \frac{x+3}{(x-1)^2 (2x-1)} dx = \frac{A}{x-1} + B \log (2x-1) + C \log (x-1) + K$ then $A+B+C=$ (A) 3 (B) 11 (C) -4 (D) -11

    ›Reveal solutionSolution

    Partial fractions give $A=-4,\ B=7,\ C=-7$, so $A+B+C=-4$.

    Set up partial fractions.

    $\frac{x+3}{(x-1)^2(2x-1)} = \frac{a}{x-1} + \frac{b}{(x-1)^2} + \frac{c}{2x-1}.$

    Clearing denominators:

    $x+3 = a(x-1)(2x-1) + b(2x-1) + c(x-1)^2.$

    Find the constants.

    • $x=1:\ 4 = b(1) \Rightarrow b = 4.$
    • $x=\tfrac12:\ \tfrac72 = c\left(-\tfrac12\right)^2 = \tfrac{c}{4} \Rightarrow c = 14.$
    • Coefficient of $x^2$: $2a + c = 0 \Rightarrow a = -7.$

    Integrate. …

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12. TG EAPCET 2022Set eng-2022-07-18-AN1 markMCQ
    Q.Let $x \neq -\dfrac{3}{5}, -\dfrac{2}{5}$, if $f\left(\dfrac{2x+1}{5x+3}\right) = x+2$, then $\int f(x)\,dx =$ (A) $\dfrac{7}{5}x - \dfrac{1}{5}\log|5x+3| + c$ (B) $\dfrac{7}{5}x - \dfrac{1}{25}\log|5x+3| + c$ (C) $\dfrac{7}{5}x - \dfrac{1}{25}\log|5x-2| + c$ (D) $\dfrac{7}{5}x - \dfrac{1}{5}\log|5x-2| + c$

    ›Reveal solutionSolution

    Put $t=\dfrac{2x+1}{5x+3}$, solve for $x$ to get $f(t)=\dfrac{7t-3}{5t-2}$. Dividing and integrating gives $\dfrac{7}{5}x-\dfrac{1}{25}\log|5x-2|+c$ — option (C).

    Let $t=\dfrac{2x+1}{5x+3}$. Then

    $t(5x+3)=2x+1\ \Rightarrow\ x(5t-2)=1-3t\ \Rightarrow\ x=\frac{1-3t}{5t-2}.$

    Since $f\!\left(\dfrac{2x+1}{5x+3}\right)=x+2$, we have

    $f(t)=x+2=\frac{1-3t}{5t-2}+2=\frac{1-3t+2(5t-2)}{5t-2}=\frac{7t-3}{5t-2}.$

    So $f(x)=\dfrac{7x-3}{5x-2}$. Divide to separate the integrable part:

    $\frac{7x-3}{5x-2}=\frac{7}{5}+\frac{-\tfrac{1}{5}}{5x-2},$ …

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