Q.Integrate the following function: $\int (2x - 3\cos x + e^x) dx$

Showing the 12 most recent of 17 on this concept.

1. TG EAPCET 2022Set eng-2022-07-20-FN1 markMCQ
   Q.If $\dfrac{3\pi}{4} < x < \dfrac{7\pi}{4}$ then $\int \left(2^x - \sqrt{1 + \sin 2x} + \dfrac{1}{x^2} - \dfrac{1}{x}\right) dx =$ (A) $\dfrac{2^x}{\log 2} - \sin x + \cos x - \dfrac{1}{x} - \log x + c$ (B) $2^x \log 2 + \sin x - \cos x - \dfrac{1}{x} + \dfrac{1}{x^2} + c$ (C) $\dfrac{2^x}{\log 2} + \sin x - \cos x - \dfrac{1}{x} - \log x + c$ (D) $2^x \log 2 - \sin x + \cos x - \dfrac{1}{x} + \dfrac{1}{x^2} + c$

   ›Reveal solutionSolution

   The key is to simplify $\sqrt{1+\sin 2x}$ using the identity $1+\sin 2x = (\sin x + \cos x)^2$, then handle the sign based on the given interval. The integral yields $\frac{2^x}{\log 2} + \sin x - \cos x - \frac{1}{x} - \log x + c$, matching option (C).

   The problem gives a definite interval for $x$: $\frac{3\pi}{4} < x < \frac{7\pi}{4}$. That interval is the entire clue — it tells you what sign to take when you simplify the square root. Without the interval, the square root would be ambiguous; with it, the sign is forced.

   Let’s break the integrand into four separate pieces and integrate each one.

   1. The term $2^x$ The integral of $a^x$ is $\frac{a^x}{\log a}$. Here $a=2$, so

   $\int 2^x \, dx = \frac{2^x}{\log 2} + c_1.$

   (In many Indian exams, $\log$ means natural log, so $\log 2 = \log 2$.)

   2. The term $-\sqrt{1+\sin 2x}$ Use the identity:

   $1 + \sin 2x = \sin^2 x + \cos^2 x + 2\sin x \cos x = (\sin x + \cos x)^2.$

   So $\sqrt{1+\sin 2x} = |\sin x + \cos x|$.

   Now, on the interval $\frac{3\pi}{4} < x < \frac{7\pi}{4}$, what is the sign of $\sin x + \cos x$?

   • At $x = \frac{3\pi}{4}$, $\sin x = \frac{1}{\sqrt{2}}$, $\cos x = -\frac{1}{\sqrt{2}}$, sum = $0$.
   • For $x$ just greater than $\frac{3\pi}{4}$, $\sin x$ is still positive but decreasing, $\cos x$ is negative and becoming more negative, so the sum is negative.
   • At $x = \pi$, $\sin \pi = 0$, $\cos \pi = -1$, sum = $-1$ (negative).
   • At $x = \frac{7\pi}{4}$, $\sin x = -\frac{1}{\sqrt{2}}$, $\cos x = \frac{1}{\sqrt{2}}$, sum = $0$ again. So throughout the open interval, $\sin x + \cos x < 0$. Hence

   $|\sin x + \cos x| = -(\sin x + \cos x).$

   Therefore

   $\int -\sqrt{1+\sin 2x} \, dx = \int -[ -(\sin x + \cos x) ] \, dx = \int (\sin x + \cos x) \, dx.$

   Integrating:

   $\int \sin x \, dx = -\cos x, \quad \int \cos x \, dx = \sin x,$

   so

   $\int (\sin x + \cos x) \, dx = -\cos x + \sin x + c_2.$ …

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2. 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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3. 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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4. 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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5. 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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6. 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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7. 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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8. 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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9. 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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10. 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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11. TG EAPCET 2022Set eng-2022-07-18-FN1 markMCQ
    Q.Let $f(x) = \int \frac{2x^3 - 3x^2 + 4x - 5}{x^2} dx$ and $f(1) = 1$. Then $f(5) =$ (A) $10 + 4 \log 5$ (B) $10 - 4 \log 5$ (C) $9 + 4 \log 5$ (D) $9 - 4 \log 5$

    ›Reveal solutionSolution

    Splitting the integrand term by term gives $f(x)=x^2-3x+4\log x+\dfrac{5}{x}+C$; using $f(1)=1$ fixes $C=-2$, so $f(5)=9+4\log 5$, option (C).

    Divide each term of the numerator by $x^2$:

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

    Integrate term by term:

    $f(x)=\int\!\left(2x-3+\frac{4}{x}-5x^{-2}\right)dx=x^2-3x+4\log|x|+\frac{5}{x}+C,$

    since $\displaystyle\int -5x^{-2}\,dx=\frac{5}{x}$.

    Fix the constant. Using $f(1)=1$: …

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12. TG EAPCET 2021Set eng-2021-08-05-AN1 markMCQ
    Q.If $\int \frac{1}{(x-2)^5 (x-1)} dx = \sum_{r=-4}^{-1} A_r \frac{(x-2)^r}{x-1} + \sum_{r=1}^{3} A_r \frac{(x-2)^r}{x-1} + B f(x)$, then $f(x) =$ (A) $\log(x-2) - \log(x-1)$ (B) $\left(\frac{x-2}{x-1}\right) + \log x$ (C) $x + \log\left(\frac{x-2}{x-1}\right)$ (D) $\log x$

    ›Reveal solutionSolution

    The integral is tackled by rewriting the integrand using a substitution that simplifies the rational function, leading to a series expansion. The final result shows that $f(x) = \log\left(\frac{x-2}{x-1}\right)$, which corresponds to option (A).

    The key idea here is that the integrand $\frac{1}{(x-2)^5 (x-1)}$ is a rational function with a repeated linear factor $(x-2)^5$ and a simple linear factor $(x-1)$. Direct partial fractions would be messy, but a clever substitution — letting $t = \frac{x-2}{x-1}$ — transforms the integral into something much simpler. Why does this work? Because the derivative of $t$ relates neatly to the denominator, and the powers of $(x-2)$ and $(x-1)$ combine into a single variable. This is a classic technique for integrals of the form $\frac{1}{(x-a)^m (x-b)^n}$.

    Let’s work through it step by step.

    1. Set up the substitution. Let $t = \frac{x-2}{x-1}$. Then $t$ is a rational function that captures the ratio of the two linear factors. Differentiate:

    $\frac{dt}{dx} = \frac{(1)(x-1) - (x-2)(1)}{(x-1)^2} = \frac{x-1 - x + 2}{(x-1)^2} = \frac{1}{(x-1)^2}.$

    So $dt = \frac{dx}{(x-1)^2}$.

    2. Rewrite the integrand in terms of $t$. Notice that $x-2 = t(x-1)$. Also, from $t = \frac{x-2}{x-1}$, we can solve for $x-1$:

    $t = \frac{x-2}{x-1} \implies t(x-1) = x-2 \implies tx - t = x-2 \implies tx - x = t - 2 \implies x(t-1) = t-2 \implies x = \frac{t-2}{t-1}.$

    Then $x-1 = \frac{t-2}{t-1} - 1 = \frac{t-2 - (t-1)}{t-1} = \frac{-1}{t-1}$. So $x-1 = \frac{1}{1-t}$ (since $-1/(t-1) = 1/(1-t)$).

    Also, $x-2 = t(x-1) = \frac{t}{1-t}$.

    3. Express the integrand and $dx$. The integrand is $\frac{1}{(x-2)^5 (x-1)}$. Substitute:

    $(x-2)^5 = \left(\frac{t}{1-t}\right)^5 = \frac{t^5}{(1-t)^5}, \quad x-1 = \frac{1}{1-t}.$

    So the product $(x-2)^5 (x-1) = \frac{t^5}{(1-t)^5} \cdot \frac{1}{1-t} = \frac{t^5}{(1-t)^6}$.

    Hence the integrand becomes $\frac{1}{(x-2)^5 (x-1)} = \frac{(1-t)^6}{t^5}$.

    Now, $dx$: from $dt = \frac{dx}{(x-1)^2}$, we have $dx = (x-1)^2 dt = \left(\frac{1}{1-t}\right)^2 dt = \frac{dt}{(1-t)^2}$.

    4. Transform the integral.

    $\int \frac{1}{(x-2)^5 (x-1)} dx = \int \frac{(1-t)^6}{t^5} \cdot \frac{dt}{(1-t)^2} = \int \frac{(1-t)^4}{t^5} dt.$

    Expand $(1-t)^4 = 1 - 4t + 6t^2 - 4t^3 + t^4$. So the integral is:

    $\int \frac{1 - 4t + 6t^2 - 4t^3 + t^4}{t^5} dt = \int \left( t^{-5} - 4t^{-4} + 6t^{-3} - 4t^{-2} + t^{-1} \right) dt.$

    5. Integrate term by term.

    $\int t^{-5} dt = \frac{t^{-4}}{-4}, \quad \int -4t^{-4} dt = -4 \cdot \frac{t^{-3}}{-3} = \frac{4}{3} t^{-3},$

    $\int 6t^{-3} dt = 6 \cdot \frac{t^{-2}}{-2} = -3 t^{-2}, \quad \int -4t^{-2} dt = -4 \cdot \frac{t^{-1}}{-1} = 4 t^{-1},$

    $\int t^{-1} dt = \log |t|.$

    So the integral is:

    $-\frac{1}{4} t^{-4} + \frac{4}{3} t^{-3} - 3 t^{-2} + 4 t^{-1} + \log |t| + C.$

    6. Substitute back $t = \frac{x-2}{x-1}$. Then $t^{-1} = \frac{x-1}{x-2}$, $t^{-2} = \left(\frac{x-1}{x-2}\right)^2$, etc. The integral becomes:

    $-\frac{1}{4} \left(\frac{x-1}{x-2}\right)^4 + \frac{4}{3} \left(\frac{x-1}{x-2}\right)^3 - 3 \left(\frac{x-1}{x-2}\right)^2 + 4 \left(\frac{x-1}{x-2}\right) + \log \left| \frac{x-2}{x-1} \right| + C.$

    7. Match with the given form. The problem states: $$$$

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