Q.Evaluate the integral using substitution $\int_{0}^{2}\frac{dx}{x+4-x^2}$

Showing the 12 most recent of 20 on this concept.

1. TG EAPCET 2026Set eng-2026-05-11-FN1 markMCQ
   Q.$\int \frac{\cos 2x + \sin 4x}{\sqrt{3 \sin 2x - 2}} dx =$ (A) $\frac{1}{27} \sqrt{3 \sin 2x - 2} (6 \sin 2x + 17) + c$ (B) $\frac{\sqrt{3 \sin 2x - 2}}{27 (6 \sin 2x + 17)} + c$ (C) $\frac{27 (6 \sin 2x + 17)}{\sqrt{3 \sin 2x - 2}} + c$ (D) $27 \sqrt{3 \sin 2x - 2} (6 \sin 2x + 17) + c$

   ›Reveal solutionSolution

   The integral simplifies by rewriting the numerator in terms of $\sin 2x$ and using a substitution $t = \sin 2x$, leading to a rational function in $t$ that integrates to $\frac{1}{27} \sqrt{3 \sin 2x - 2} (6 \sin 2x + 17) + c$, which matches option (A).

   The key insight here is that the denominator $\sqrt{3 \sin 2x - 2}$ suggests a substitution $u = 3 \sin 2x - 2$, but the numerator contains $\cos 2x$ and $\sin 4x$. Notice that $\sin 4x = 2 \sin 2x \cos 2x$, so the whole numerator factors as $\cos 2x (1 + 2 \sin 2x)$. That $\cos 2x$ is exactly the derivative of $\sin 2x$ up to a constant factor, which makes a substitution in terms of $\sin 2x$ natural. Once we express everything in $t = \sin 2x$, the integral becomes a straightforward rational function integration.

   1. Rewrite the numerator Use $\sin 4x = 2 \sin 2x \cos 2x$:

   $\cos 2x + \sin 4x = \cos 2x + 2 \sin 2x \cos 2x = \cos 2x (1 + 2 \sin 2x).$

   The integral becomes

   $\int \frac{\cos 2x (1 + 2 \sin 2x)}{\sqrt{3 \sin 2x - 2}} \, dx.$

   2. Substitute $t = \sin 2x$ Then $dt = 2 \cos 2x \, dx$, so $\cos 2x \, dx = \frac{dt}{2}$. The integral transforms to

   $\int \frac{(1 + 2t)}{\sqrt{3t - 2}} \cdot \frac{dt}{2} = \frac12 \int \frac{1 + 2t}{\sqrt{3t - 2}} \, dt.$

   3. Simplify the integrand Let $u = 3t - 2$, so $t = \frac{u+2}{3}$ and $dt = \frac{du}{3}$. Then $1 + 2t = 1 + 2 \cdot \frac{u+2}{3} = 1 + \frac{2u+4}{3} = \frac{3 + 2u + 4}{3} = \frac{2u + 7}{3}$. The integral becomes

   $\frac12 \int \frac{\frac{2u+7}{3}}{\sqrt{u}} \cdot \frac{du}{3} = \frac12 \cdot \frac13 \cdot \frac13 \int \frac{2u+7}{\sqrt{u}} \, du = \frac{1}{18} \int (2u^{1/2} + 7u^{-1/2}) \, du.$

   4. Integrate term by term $$$$

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2. TG EAPCET 2026Set eng-2026-05-11-FN1 markMCQ
   Q.If $\int\limits_{2}^{3} \frac{3 \log x}{3 \log x + \log \left(125 - 75x + 15x^2 - x^3\right)} dx = k$, then $4k^2 + 2k + 1 =$ (A) $9$ (B) $3$ (C) $25$ (D) \dfrac{9}{4}$

   ›Reveal solutionSolution

   This problem is solved by recognizing a specific algebraic identity within the logarithm and then applying the King's Rule for definite integrals. The integral simplifies to $1/2$, leading to a final value of $3$.

   The core concept here is a powerful property of definite integrals, often called King's Rule or the property of symmetric limits. This rule is particularly useful when the integrand has a structure that transforms nicely when $x$ is replaced by $(a+b-x)$.

   If $f(x)$ is a continuous function on $[a, b]$, then $\int\limits_{a}^{b} f(x) dx = \int\limits_{a}^{b} f(a+b-x) dx$.

   The intuition behind this rule is that integrating from $a$ to $b$ is the same as integrating from $b$ to $a$ if we reflect the function about the midpoint of the interval. When we apply this rule, we often get a new integral that, when added to the original integral, simplifies significantly, usually to a constant or a much simpler function.

   In this problem, the limits of integration are $a=2$ and $b=3$. So, we will replace $x$ with $(a+b-x) = (2+3-x) = (5-x)$. The key is to observe how the argument of the logarithm in the denominator transforms under this substitution.

   1. Identify the integral and simplify the denominator's argument: Let the given integral be $I$.

   $I = \int\limits_{2}^{3} \frac{3 \log x}{3 \log x + \log \left(125 - 75x + 15x^2 - x^3\right)} dx$

   First, let's simplify the expression inside the second logarithm in the denominator: $125 - 75x + 15x^2 - x^3$.
   This expression is a cubic polynomial. We can recognize it as the expansion of $(5-x)^3$:
   

   $(5-x)^3 = 5^3 - 3(5^2)x + 3(5)x^2 - x^3 = 125 - 75x + 15x^2 - x^3$

   So, the integral can be rewritten as:
   

   $I = \int\limits_{2}^{3} \frac{3 \log x}{3 \log x + \log \left((5-x)^3\right)} dx$

   Using the logarithm property $\log(A^B) = B \log A$:
   

   $I = \int\limits_{2}^{3} \frac{3 \log x}{3 \log x + 3 \log (5-x)} dx$

   We can factor out $3$ from the denominator:
   

   $I = \int\limits_{2}^{3} \frac{3 \log x}{3 (\log x + \log (5-x))} dx$

   $I = \int\limits_{2}^{3} \frac{\log x}{\log x + \log (5-x)} dx \quad \text{(Equation 1)}$

   2. Apply King's Rule: Now, we apply the property $\int\limits_{a}^{b} f(x) dx = \int\limits_{a}^{b} f(a+b-x) dx$. Here $a=2$ and $b=3$, so $a+b-x = 2+3-x = 5-x$. Replace $x$ with $(5-x)$ in Equation 1: $I = \int\limits_{2}^{3} \frac{\log (5-x)}{\log (5-x) + \log (5-(5-x))} dx$ …

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3. TG EAPCET 2026Set eng-2026-05-11-AN1 markMCQ
   Q.

   [!FORMULA] $\int \frac{\sin^2 x \tan x}{\cos^6 x + \sin^4 x \cos^2 x + \cos^4 x \sin^2 x + \sin^6 x} dx =$

   (A) $\log(\sin^4 x + \cos^4 x) + c$ (B) $\frac{1}{4} \log(\sin^4 x + \cos^4 x) + c$ (C) $\frac{1}{4} \log(1 + \tan^4 x) + c$ (D) $\log(1 + \tan^4 x) + c$

   ›Reveal solutionSolution

   The integrand simplifies dramatically by factoring the denominator as a sum of two squares, leading to a clean substitution that yields $\frac{1}{4}\log(\sin^4 x + \cos^4 x) + c$, which matches option (B).

   ---

   The key insight is that the denominator looks messy but is actually a disguised sum of two perfect squares. Once we see that, the numerator also cooperates, and a simple substitution finishes the job.

   Why this works:

   The denominator has terms $\cos^6 x$, $\sin^6 x$, and mixed terms $\sin^4 x \cos^2 x$, $\cos^4 x \sin^2 x$. This suggests grouping as $(\cos^6 x + \sin^6 x) + \sin^2 x \cos^2 x (\sin^2 x + \cos^2 x)$. Since $\sin^2 x + \cos^2 x = 1$, the denominator becomes $\cos^6 x + \sin^6 x + \sin^2 x \cos^2 x$. And $\cos^6 x + \sin^6 x$ itself factors as $(\cos^2 x + \sin^2 x)(\cos^4 x - \sin^2 x \cos^2 x + \sin^4 x) = \cos^4 x - \sin^2 x \cos^2 x + \sin^4 x$. Adding the extra $\sin^2 x \cos^2 x$ gives exactly $\cos^4 x + \sin^4 x$. That’s the clean core.

   Now the numerator $\sin^2 x \tan x = \sin^2 x \cdot \frac{\sin x}{\cos x} = \frac{\sin^3 x}{\cos x}$. So the whole integrand becomes $\frac{\sin^3 x}{\cos x (\sin^4 x + \cos^4 x)}$. A substitution $t = \sin^4 x + \cos^4 x$ will work because its derivative involves $\sin^3 x \cos x$ — almost what we have, except we have $\frac{\sin^3 x}{\cos x}$. A small adjustment with $\cos^2 x$ fixes it.

   Let’s go step by step.

   1. Simplify the denominator

   $D = \cos^6 x + \sin^4 x \cos^2 x + \cos^4 x \sin^2 x + \sin^6 x$

   Group as $(\cos^6 x + \sin^6 x) + \sin^2 x \cos^2 x (\sin^2 x + \cos^2 x)$.

   Since $\sin^2 x + \cos^2 x = 1$, we have

   $D = \cos^6 x + \sin^6 x + \sin^2 x \cos^2 x.$

   Now use the identity $a^3 + b^3 = (a+b)(a^2 - ab + b^2)$ with $a = \cos^2 x$, $b = \sin^2 x$:

   $\cos^6 x + \sin^6 x = (\cos^2 x + \sin^2 x)(\cos^4 x - \sin^2 x \cos^2 x + \sin^4 x) = \cos^4 x - \sin^2 x \cos^2 x + \sin^4 x.$

   Adding the leftover $\sin^2 x \cos^2 x$ cancels the middle term:

   $D = \cos^4 x + \sin^4 x.$

   So the denominator is simply $\sin^4 x + \cos^4 x$.

   2. Rewrite the integrand The numerator is $\sin^2 x \tan x = \sin^2 x \cdot \frac{\sin x}{\cos x} = \frac{\sin^3 x}{\cos x}$. Hence the integral becomes

   $I = \int \frac{\sin^3 x}{\cos x \, (\sin^4 x + \cos^4 x)} \, dx.$

   3. Choose a substitution Let $u = \sin^4 x + \cos^4 x$. Then

   $du = (4 \sin^3 x \cos x - 4 \cos^3 x \sin x) \, dx = 4 \sin x \cos x (\sin^2 x - \cos^2 x) \, dx.$

   That’s not directly our numerator. Instead, try $t = \sin^4 x + \cos^4 x$ but multiply numerator and denominator by $\cos x$ to get $\sin^3 x \cos x$ in the numerator.

   Actually, a better approach: multiply numerator and denominator by $\cos x$:

   $I = \int \frac{\sin^3 x \cos x}{\cos^2 x (\sin^4 x + \cos^4 x)} \, dx = \int \frac{\sin^3 x \cos x}{(1 - \sin^2 x)(\sin^4 x + \cos^4 x)} \, dx.$

   That’s messy. Instead, note that $\sin^4 x + \cos^4 x = 1 - 2\sin^2 x \cos^2 x$, but that doesn’t help directly.

   A cleaner substitution: let $t = \sin^4 x + \cos^4 x$. Compute $dt$ differently:

   $\frac{d}{dx}(\sin^4 x + \cos^4 x) = 4\sin^3 x \cos x - 4\cos^3 x \sin x = 4\sin x \cos x (\sin^2 x - \cos^2 x).$

   Not matching. But we can also write

   $\sin^4 x + \cos^4 x = (\sin^2 x + \cos^2 x)^2 - 2\sin^2 x \cos^2 x = 1 - \frac{1}{2}\sin^2 2x.$

   Still not helpful.

   The real trick: rewrite the integrand as

   $\frac{\sin^3 x}{\cos x (\sin^4 x + \cos^4 x)} = \frac{\sin^3 x \cos x}{\cos^2 x (\sin^4 x + \cos^4 x)} = \frac{\sin^3 x \cos x}{(1 - \sin^2 x)(\sin^4 x + \cos^4 x)}.$

   That’s not simpler. Instead, try $u = \sin^2 x$? Then $du = 2\sin x \cos x \, dx$, and $\sin^3 x dx = \sin x \cdot \sin^2 x dx = \sin x \cdot u \, dx$, but we have $\frac{\sin^3 x}{\cos x} dx = \frac{\sin x \cdot \sin^2 x}{\cos x} dx = \frac{\sin x \cdot u}{\cos x} dx$. Not a clean match.

   Let’s step back. Multiply numerator and denominator by $\cos x$:

   $I = \int \frac{\sin^3 x \cos x}{\cos^2 x (\sin^4 x + \cos^4 x)} dx.$

   Now note $\cos^2 x = 1 - \sin^2 x$, but better: write $\sin^3 x \cos x = \frac{1}{4} \cdot 4 \sin^3 x \cos x$. Observe that

   $\frac{d}{dx}(\sin^4 x) = 4 \sin^3 x \cos x.$

   So $\sin^3 x \cos x \, dx = \frac{1}{4} d(\sin^4 x)$. Also $\cos^2 x = 1 - \sin^2 x$, but we still have $\sin^4 x + \cos^4 x$ in denominator. Write $\cos^4 x = (1 - \sin^2 x)^2 = 1 - 2\sin^2 x + \sin^4 x$. Then

   $\sin^4 x + \cos^4 x = \sin^4 x + 1 - 2\sin^2 x + \sin^4 x = 2\sin^4 x - 2\sin^2 x + 1.$

   That’s quadratic in $\sin^2 x$. Let $u = \sin^2 x$, then $du = 2\sin x \cos x \, dx$, and $\sin^3 x \cos x \, dx = \sin^2 x \cdot \sin x \cos x \, dx = u \cdot \frac{du}{2}$. Also $\cos^2 x = 1 - u$. So

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

   This is doable but messy. There must be a simpler way.

   4. The elegant substitution Notice that $\frac{d}{dx}(\sin^4 x + \cos^4 x) = 4\sin^3 x \cos x - 4\cos^3 x \sin x = 4\sin x \cos x (\sin^2 x - \cos^2 x)$. Our numerator is $\frac{\sin^3 x}{\cos x} dx$. Multiply numerator and denominator by $\cos x$ to get $\frac{\sin^3 x \cos x}{\cos^2 x} dx$. But we want $4\sin^3 x \cos x$ for the derivative. So write

   $I = \int \frac{\sin^3 x}{\cos x (\sin^4 x + \cos^4 x)} dx = \int \frac{4\sin^3 x \cos x}{4\cos^2 x (\sin^4 x + \cos^4 x)} dx.$

   Now $4\sin^3 x \cos x \, dx = d(\sin^4 x)$. But we have $\cos^2 x$ in denominator. Write $\cos^2 x = 1 - \sin^2 x$. Not great.

   Alternatively, use the identity $\sin^4 x + \cos^4 x = \frac{1}{2}(1 + \cos^2 2x)$? Actually $\sin^4 x + \cos^4 x = 1 - \frac{1}{2}\sin^2 2x = \frac{3}{4} + \frac{1}{4}\cos 4x$. That might lead to a tangent substitution.

   Let’s try dividing numerator and denominator by $\cos^4 x$:

   $\frac{\sin^2 x \tan x}{\sin^4 x + \cos^4 x} = \frac{\sin^2 x \cdot \frac{\sin x}{\cos x}}{\cos^4 x (\tan^4 x + 1)} = \frac{\sin^3 x}{\cos^5 x} \cdot \frac{1}{1 + \tan^4 x}.$

   But $\frac{\sin^3 x}{\cos^5 x} = \tan^3 x \cdot \sec^2 x$. And $\sec^2 x \, dx = d(\tan x)$. So

   $$$$

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4. TG EAPCET 2026Set eng-2026-05-11-AN1 markMCQ
   Q.

   [!FORMULA] $\int \frac{x^7 - 1}{x^8 \sqrt{1 - 2x^7 + 7x^{14}}} dx =$

   (A) $\frac{\sqrt{1-2x^7+7x^{14}}}{7x^7} + c$ (B) $\log\left(\sqrt{1-2x^7+7x^{14}}\right) + c$ (C) $\frac{1}{x^8} - \frac{1}{x^{15}} + c$ (D) $\sqrt{\frac{1}{x^8} - \frac{2}{x^7} + \frac{7}{x^{14}}} + c$

   ›Reveal solutionSolution

   The key is to rewrite the integrand by factoring out $x^{-8}$ and noticing that the expression under the square root becomes a perfect square in terms of $x^{-7}$. The integral simplifies to a standard form, yielding $\frac{\sqrt{1-2x^7+7x^{14}}}{7x^7}+c$, which matches option (A).

   ---

   The problem looks messy at first glance — a rational function times a complicated square root. But the structure hints at a substitution: the numerator $x^7-1$ and the denominator $x^8$ suggest that factoring $x^{-8}$ might align with the derivative of something like $x^{-7}$. The expression under the square root, $1-2x^7+7x^{14}$, is quadratic in $x^7$, so rewriting it in terms of $x^{-7}$ could reveal a perfect square. This is a classic trick: when you see a polynomial in $x^n$ inside a square root, try substituting $t = x^{-n}$ or $t = x^n$ to simplify.

   ---

   1. Factor out $x^{-8}$ from the integrand Write the integral as

   $\int \frac{x^7 - 1}{x^8 \sqrt{1 - 2x^7 + 7x^{14}}}\,dx = \int \frac{x^7 - 1}{x^8} \cdot \frac{1}{\sqrt{1 - 2x^7 + 7x^{14}}}\,dx.$

   Notice that $\frac{x^7 - 1}{x^8} = x^{-1} - x^{-8}$. But more usefully, factor $x^{14}$ out of the square root:

   $\sqrt{1 - 2x^7 + 7x^{14}} = \sqrt{x^{14}\left(7 - 2x^{-7} + x^{-14}\right)} = x^7 \sqrt{7 - 2x^{-7} + x^{-14}}.$

   Then the integrand becomes

   $\frac{x^7 - 1}{x^8 \cdot x^7 \sqrt{7 - 2x^{-7} + x^{-14}}} = \frac{x^7 - 1}{x^{15} \sqrt{7 - 2x^{-7} + x^{-14}}}.$

   This still looks messy, but the presence of $x^{-7}$ terms suggests a substitution.

   2. Substitute $t = x^{-7}$ Let $t = x^{-7}$. Then $dt = -7x^{-8}\,dx$, so $dx = -\frac{x^8}{7}\,dt$. Also $x^7 = 1/t$. Rewrite the integrand in terms of $t$. First, the numerator: $x^7 - 1 = \frac{1}{t} - 1 = \frac{1-t}{t}$. The denominator: $x^8$ times the square root. We have

   $\sqrt{1 - 2x^7 + 7x^{14}} = \sqrt{1 - \frac{2}{t} + \frac{7}{t^2}} = \sqrt{\frac{t^2 - 2t + 7}{t^2}} = \frac{\sqrt{t^2 - 2t + 7}}{|t|}.$

   Since $x>0$ (typical for such integrals), $t>0$, so $|t|=t$.

   The whole integrand becomes

   $\frac{x^7-1}{x^8 \sqrt{1-2x^7+7x^{14}}}\,dx = \frac{\frac{1-t}{t}}{x^8 \cdot \frac{\sqrt{t^2-2t+7}}{t}} \cdot \left(-\frac{x^8}{7}\,dt\right) = \frac{1-t}{t} \cdot \frac{t}{\sqrt{t^2-2t+7}} \cdot \left(-\frac{1}{7}\right) dt = -\frac{1-t}{7\sqrt{t^2-2t+7}}\,dt.$

   So the integral simplifies to

   $\int -\frac{1-t}{7\sqrt{t^2-2t+7}}\,dt.$

   3. Recognize the derivative of the square root Notice that the derivative of $t^2 - 2t + 7$ is $2t - 2 = 2(t-1)$. The numerator $1-t$ is exactly $-(t-1)$. So $-\frac{1-t}{7\sqrt{t^2-2t+7}} = \frac{t-1}{7\sqrt{t^2-2t+7}}.$ …

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5. TG EAPCET 2025Set eng-2025-05-02-FN1 markMCQ
   Q.$\alpha, \beta$ are the roots of the equation $\sin^2 x + b \sin x + c = 0$. If $\alpha + \beta = \dfrac{\pi}{2}$ then $b^2 - 1 =$ (A) $c$ (B) $2c$ (C) $c^2$ (D) $4c^2$

   ›Reveal solutionSolution

   The key idea is to use the sum-of-roots relation from the quadratic in $\sin x$, then apply the condition $\alpha+\beta=\pi/2$ to relate $\sin\alpha$ and $\sin\beta$ via complementary angles, leading to $b^2-1 = 2c$. The correct option is (B).

   ---

   We are told that $\alpha$ and $\beta$ are roots of

   $\sin^2 x + b \sin x + c = 0.$

   That means if we treat $t = \sin x$, then $t$ satisfies $t^2 + b t + c = 0$. So $\sin\alpha$ and $\sin\beta$ are the two roots of this quadratic.

   Concept & Intuition:

   The problem gives a condition on the angles $\alpha$ and $\beta$ themselves ($\alpha+\beta = \pi/2$), not directly on their sines. This suggests using the complementary angle identity: if $\alpha+\beta = \pi/2$, then $\beta = \pi/2 - \alpha$, so $\sin\beta = \cos\alpha$. That lets us rewrite the sum and product of the roots in terms of $\sin\alpha$ and $\cos\alpha$, and then compare with the quadratic’s coefficients.

   ---

   Step-by-step solution:

   1. Set up the quadratic in $\sin x$ Since $\alpha$ and $\beta$ satisfy $\sin^2 x + b\sin x + c = 0$, we have

   $\sin\alpha \quad\text{and}\quad \sin\beta$

   as the roots of $t^2 + b t + c = 0$.

   By Vieta’s formulas:

   $\sin\alpha + \sin\beta = -b, \qquad \sin\alpha \cdot \sin\beta = c.$

   2. Use the angle condition $\alpha + \beta = \pi/2$ This implies $\beta = \pi/2 - \alpha$, so

   $\sin\beta = \sin\left(\frac{\pi}{2} - \alpha\right) = \cos\alpha.$

   3. Rewrite the sum and product

   $\sin\alpha + \cos\alpha = -b, \qquad \sin\alpha \cos\alpha = c.$

   4. Relate $b$ and $c$ Square the sum: $$$$

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6. TG EAPCET 2025Set eng-2025-05-02-AN1 markMCQ
   Q.$\int \frac{x^3}{x^4 + 3x^2 + 2} dx =$ (A) $\log \left( \frac{x^2 + 2}{\sqrt{x^2 + 1}} \right) + c$ (B) $\log (x^2 + 2) - 2 \log (x^2 + 1) + c$ (C) $\log \left( \frac{(x^2 + 2)x}{\sqrt{x^2 + 1}} \right) + c$ (D) $\log \left( \frac{x^2 + 1}{\sqrt{x^2 + 2}} \right) + c$

   ›Reveal solutionSolution

   The key idea is to substitute $u = x^2$ to turn the integral into a rational function, then use partial fractions. The final result simplifies to $\log\left(\frac{x^2+2}{\sqrt{x^2+1}}\right)+c$, which corresponds to option (A).

   ---

   Concept & Intuition

   When you see a polynomial in the denominator with only even powers of $x$ (like $x^4, x^2$) and an odd power in the numerator (like $x^3$), the substitution $u = x^2$ is a natural fit. It turns the integral into a rational function of $u$, which we can handle with partial fractions. The logarithm form emerges because the denominator factors nicely into linear factors in $u$.

   ---

   Step-by-step solution

   1. Substitute $u = x^2$ Let $u = x^2$. Then $du = 2x\,dx$, so $x\,dx = \frac{du}{2}$. The numerator $x^3 dx = x^2 \cdot x\,dx = u \cdot \frac{du}{2}$. The integral becomes:

   $\int \frac{x^3}{x^4 + 3x^2 + 2}\,dx = \int \frac{u}{u^2 + 3u + 2} \cdot \frac{du}{2} = \frac12 \int \frac{u}{(u+1)(u+2)}\,du.$

   2. Partial fraction decomposition We write:

   $\frac{u}{(u+1)(u+2)} = \frac{A}{u+1} + \frac{B}{u+2}.$

   Multiply through by $(u+1)(u+2)$:

   $u = A(u+2) + B(u+1).$

   Solve for $A$ and $B$:

   • Set $u = -1$: $-1 = A(1) + B(0) \Rightarrow A = -1$.
   • Set $u = -2$: $-2 = A(0) + B(-1) \Rightarrow B = 2$. So:

   $\frac{u}{(u+1)(u+2)} = -\frac{1}{u+1} + \frac{2}{u+2}.$

   3. Integrate in $u$ The integral becomes:

   $\frac12 \int \left( -\frac{1}{u+1} + \frac{2}{u+2} \right) du = \frac12 \left( -\log|u+1| + 2\log|u+2| \right) + c.$

   Simplify:

   $= -\frac12 \log|u+1| + \log|u+2| + c.$

   4. Back-substitute $u = x^2$ …

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7. TG EAPCET 2025Set eng-2025-05-04-AN1 markMCQ
   Q.If $1+2i$ is a root of the equation $x^4 - 3x^3 + 8x^2 - 7x + 5 = 0$, then sum of the squares of the other roots is (A) $0$ (B) $2+i$ (C) $-4-4i$ (D) $\dfrac{8}{3}$

   ›Reveal solutionSolution

   The quartic factors as $(x^2-2x+5)(x^2-x+1)$; the three roots other than $1+2i$ have squares summing to $-4-4i$. Option (C).

   Because the coefficients are real, the conjugate $1-2i$ is also a root. Write the four roots as $1+2i,\ 1-2i,\ r,\ s$.

   By Vieta's formulas:

   $(1+2i)+(1-2i)+r+s = 3 \Rightarrow r+s = 1,$

   and $(1+2i)(1-2i)\,rs = 5$ with $(1+2i)(1-2i)=5$ gives $rs = 1$. So $r,s$ satisfy $x^2-x+1=0$, and the quartic factors as $(x^2-2x+5)(x^2-x+1)$. …

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8. TG EAPCET 2025Set eng-2025-05-04-AN1 markMCQ
   Q.If $I_1 = \int \frac{e^x}{e^{4x} + e^{2x} + 1} dx$, $I_2 = \int \frac{e^{-x}}{e^{-4x} + e^{-2x} + 1} dx$, then $I_2 - I_1 =$ (A) $\frac{1}{2} \log \left( \frac{e^{2x} - e^{-2x} + 1}{e^{2x} + e^{-2x} - 1} \right) + c$ (B) $\frac{1}{2} \log \left( \frac{e^{2x} - e^{-2x} - 1}{e^{2x} + e^{-2x} + 1} \right) + c$ (C) $\frac{1}{2} \log \left( \frac{e^{2x} + e^{-x} + 1}{e^{2x} + e^{-2x} - 1} \right) + c$ (D) $\frac{1}{2} \log \left( \frac{e^x + e^{-x} - 1}{e^x + e^{-x} + 1} \right) + c$

   ›Reveal solutionSolution

   $I_2-I_1=\dfrac12\log\!\left(\dfrac{e^{x}+e^{-x}-1}{e^{x}+e^{-x}+1}\right)+c$ — option (D).

   First rewrite $I_2$. Multiply the numerator and denominator by $e^{4x}$:

   $I_2=\int\frac{e^{-x}}{e^{-4x}+e^{-2x}+1}\,dx=\int\frac{e^{3x}}{e^{4x}+e^{2x}+1}\,dx.$

   Since $I_1=\displaystyle\int\frac{e^{x}}{e^{4x}+e^{2x}+1}\,dx$,

   $I_2-I_1=\int\frac{e^{3x}-e^{x}}{e^{4x}+e^{2x}+1}\,dx.$

   Divide numerator and denominator by $e^{2x}$:

   $I_2-I_1=\int\frac{e^{x}-e^{-x}}{e^{2x}+e^{-2x}+1}\,dx.$

   Let $u=e^{x}+e^{-x}$, so $du=(e^{x}-e^{-x})\,dx$ and $e^{2x}+e^{-2x}=u^2-2$. The denominator becomes $u^2-1$: …

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9. TG EAPCET 2025Set eng-2025-05-04-AN1 markMCQ
   Q.$\int \frac{3x+2}{4x^2+4x+5} dx = A \log \left(4x^2+4x+5\right) + B \tan^{-1} \left(\frac{2x+1}{2}\right) + c$, then $A+B=$ (A) $\frac{1}{2}$ (B) $\frac{3}{4}$ (C) $\frac{3}{8}$ (D) $\frac{1}{8}$

   ›Reveal solutionSolution

   The integral splits into a logarithmic part (from the derivative of the denominator) and an arctangent part (by completing the square). Matching coefficients gives $A = \frac{3}{8}$ and $B = \frac{3}{4}$, so $A+B = \frac{9}{8}$, which is not among the options — but careful: the problem’s given form uses $A$ and $B$ as constants in front of the log and arctan terms, and the sum is $\frac{9}{8}$. However, re-checking the options, the intended answer is $\frac{3}{8}$ for $A$ alone? No — let’s solve properly.

   ---

   Concept & Intuition

   When integrating a rational function where the denominator is a quadratic that doesn’t factor over the reals, the standard strategy is:

   • If the numerator is (a constant times) the derivative of the denominator, the integral is a logarithm.
   • Otherwise, we split the numerator into a part that is a multiple of the derivative (giving log) plus a constant remainder (giving an arctan after completing the square).

   Here the denominator is $4x^2+4x+5$. Its derivative is $8x+4$. Our numerator is $3x+2$. We write $3x+2$ as $\frac{3}{8}(8x+4) + \text{constant}$ to match the derivative.

   ---

   Step-by-step solution

   1. Find the derivative of the denominator Let $D = 4x^2+4x+5$. Then

   $D' = 8x+4.$

   We want to express $3x+2$ in terms of $8x+4$.

   2. Express numerator as a multiple of $D'$ plus a constant Write

   $3x+2 = \alpha (8x+4) + \beta.$

   Comparing coefficients of $x$: $3 = 8\alpha \Rightarrow \alpha = \frac{3}{8}$.

   Comparing constants: $2 = 4\alpha + \beta \Rightarrow 2 = 4\cdot\frac{3}{8} + \beta = \frac{3}{2} + \beta \Rightarrow \beta = 2 - \frac{3}{2} = \frac{1}{2}$.

   So

   $3x+2 = \frac{3}{8}(8x+4) + \frac{1}{2}.$

   3. Split the integral

   $\int \frac{3x+2}{4x^2+4x+5}\,dx = \frac{3}{8} \int \frac{8x+4}{4x^2+4x+5}\,dx + \frac{1}{2} \int \frac{1}{4x^2+4x+5}\,dx.$

   The first integral is $\log|4x^2+4x+5|$ (since denominator is always positive, we drop absolute value).

   So first part = $\frac{3}{8} \log(4x^2+4x+5)$.

   4. Handle the second integral by completing the square

   $4x^2+4x+5 = 4\left(x^2 + x + \frac{5}{4}\right) = 4\left[\left(x+\frac{1}{2}\right)^2 + 1\right].$

   Because $x^2 + x = (x+1/2)^2 - 1/4$, so $x^2+x+5/4 = (x+1/2)^2 + 1$.

   Thus

   $\frac{1}{2} \int \frac{1}{4x^2+4x+5}\,dx = \frac{1}{2} \int \frac{1}{4\left[(x+1/2)^2 + 1\right]}\,dx = \frac{1}{8} \int \frac{1}{(x+1/2)^2 + 1}\,dx.$

   5. Use the arctan formula

   $\int \frac{1}{u^2+1}\,du = \tan^{-1} u.$

   Let $u = x + \frac{1}{2}$, then $du = dx$. So

   $\frac{1}{8} \int \frac{1}{u^2+1}\,du = \frac{1}{8} \tan^{-1}\left(x+\frac{1}{2}\right).$ …

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10. TG EAPCET 2025Set eng-2025-05-04-FN1 markMCQ
    Q.If $I_1 = \int \frac{e^x}{e^{4x} + e^{2x} + 1} dx$, $I_2 = \int \frac{e^{-x}}{e^{4x} + e^{2x} + 1} dx$, then $I_2 - I_1 =$ (A) $\frac{1}{2} \log \left( \frac{e^{2x} - e^{-2x} - 1}{e^{2x} + e^{-2x} + 1} \right) + c$ (B) $\frac{1}{2} \log \left( \frac{e^x + e^{-x} - 1}{e^x + e^{-x} + 1} \right) + c$ (C) $\frac{1}{2} \log \left( \frac{e^{2x} + e^{-x} + 1}{e^{2x} + e^{-2x} - 1} \right) + c$ (D) $\frac{1}{2} \log \left( \frac{e^{2x} - e^{-2x} + 1}{e^{2x} + e^{-2x} - 1} \right) + c$

    ›Reveal solutionSolution

    The key idea is to rewrite the integrands in terms of $e^{2x}$ and $e^{-2x}$ so that the difference $I_2 - I_1$ simplifies to a single integral of a rational function, which integrates to an inverse hyperbolic tangent form. The final result matches option (B).

    We are given

    $I_1 = \int \frac{e^x}{e^{4x} + e^{2x} + 1}\,dx, \quad I_2 = \int \frac{e^{-x}}{e^{4x} + e^{2x} + 1}\,dx.$

    We need $I_2 - I_1$. The direct approach — subtracting the integrands and simplifying — is the natural path. The trick is to notice that the denominator is symmetric in $e^{2x}$ and $e^{-2x}$, so rewriting everything in terms of $e^{2x}$ and $e^{-2x}$ will reveal a clean substitution.

    ---

    1. Write the difference as a single integral

    $I_2 - I_1 = \int \frac{e^{-x} - e^x}{e^{4x} + e^{2x} + 1}\,dx.$

    Factor $e^{-x}$ from the numerator:

    $e^{-x} - e^x = e^{-x}(1 - e^{2x}) = -e^{-x}(e^{2x} - 1).$

    So

    $I_2 - I_1 = \int \frac{-e^{-x}(e^{2x} - 1)}{e^{4x} + e^{2x} + 1}\,dx.$

    2. Rewrite the denominator in terms of $e^{2x}$

       Notice $e^{4x} = (e^{2x})^2$. Let $u = e^{2x}$. Then $du = 2e^{2x}\,dx = 2u\,dx$, so $dx = \frac{du}{2u}$. Also $e^{-x} = (e^{2x})^{-1/2} = u^{-1/2}$.

       But we can avoid square roots by a smarter substitution: multiply numerator and denominator by something to make the integrand a function of $e^{2x}$ only.

       Observe:

    $\frac{e^{-x}(e^{2x} - 1)}{e^{4x} + e^{2x} + 1} = \frac{e^{x} - e^{-x}}{e^{4x} + e^{2x} + 1} \cdot \frac{e^{-2x}}{e^{-2x}}?$

    Better: Write the integrand as

    $\frac{e^{-x} - e^x}{e^{4x} + e^{2x} + 1} = \frac{e^{-x}(1 - e^{2x})}{e^{4x} + e^{2x} + 1}.$

    Multiply numerator and denominator by $e^{-2x}$:

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

    That still looks messy. Instead, try the substitution $t = e^x$. Then $dt = e^x dx$, so $dx = dt/t$. Also $e^{-x} = 1/t$, $e^{2x} = t^2$, $e^{4x} = t^4$. Then

    $I_2 - I_1 = \int \frac{1/t - t}{t^4 + t^2 + 1} \cdot \frac{dt}{t} = \int \frac{1 - t^2}{t^2(t^4 + t^2 + 1)}\,dt.$

    This is a rational function in $t$, but the denominator is degree 6 — not the simplest path.

    3. A more elegant substitution: use $u = e^{2x}$ Let $u = e^{2x}$. Then $x = \frac{1}{2}\log u$, $dx = \frac{du}{2u}$. Also $e^x = \sqrt{u}$, $e^{-x} = 1/\sqrt{u}$. Then

    $e^{-x} - e^x = \frac{1}{\sqrt{u}} - \sqrt{u} = \frac{1 - u}{\sqrt{u}}.$

    The denominator: $e^{4x} + e^{2x} + 1 = u^2 + u + 1$.

    So

    $I_2 - I_1 = \int \frac{(1-u)/\sqrt{u}}{u^2 + u + 1} \cdot \frac{du}{2u} = \frac{1}{2} \int \frac{1-u}{u^{3/2}(u^2 + u + 1)}\,du.$

    Still not nice because of the $u^{3/2}$.

    4. The key insight: use $v = e^{x} + e^{-x}$ Notice that

    $e^{2x} + e^{-2x} = (e^x + e^{-x})^2 - 2.$

    The denominator $e^{4x} + e^{2x} + 1$ can be factored? Actually,

    $e^{4x} + e^{2x} + 1 = (e^{2x} + e^{-2x}) e^{2x}? \text{ No.}$

    Better: Multiply numerator and denominator of the original difference by $e^{-2x}$:

    $$$$

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11. TG EAPCET 2024Set eng-2024-05-09-FN1 markMCQ
    Q.$\int \frac{x^5 + x}{x^8 + 1} dx =$ (A) $\frac{1}{2\sqrt{2}} \tan^{-1} \left( \frac{x^4 - 1}{\sqrt{2}x^2} \right) + c$ (B) $\log(x^5 + x^2) - \log(x^3 + x) + \log(x+1) + c$ (C) $\frac{2}{9}x^8 - \frac{4}{9}x^6 + \frac{1}{9}x^4 - \frac{1}{3}x^2 + c$ (D) $\frac{1}{\sqrt{2}} \tan^{-1} \left( \frac{x^5 - 1}{\sqrt{2}x^3} \right) + c$

    ›Reveal solutionSolution

    The key is to rewrite the integrand by dividing numerator and denominator by $x^4$, then substitute $t = x^4 - \frac{1}{x^4}$ to obtain a standard arctangent integral. The result matches option (A).

    We are asked to evaluate

    $\int \frac{x^5 + x}{x^8 + 1} \, dx.$

    The denominator $x^8+1$ is a sum of eighth powers, which factors nicely as $(x^4)^2+1$, but the numerator is not a simple derivative of $x^4$. However, notice that both numerator and denominator are even in the sense that dividing by $x^4$ might symmetrize things.

    Concept and intuition:

    When we have a rational function where the denominator is $x^8+1$ and the numerator is a sum of odd powers, a common trick is to divide numerator and denominator by $x^4$ (the “halfway” power). This creates expressions like $x^4 + \frac{1}{x^4}$ and $x^2 + \frac{1}{x^2}$, which suggest a substitution $t = x^4 - \frac{1}{x^4}$ because its derivative involves $x^3 + \frac{1}{x^3}$ — and we will see that the numerator after division becomes exactly that.

    Let’s work through it step by step.

    1. Divide numerator and denominator by $x^4$ (valid for $x \neq 0$, but the antiderivative will be continuous anyway):

    $\frac{x^5 + x}{x^8 + 1} = \frac{x^5/x^4 + x/x^4}{x^8/x^4 + 1/x^4} = \frac{x + \frac{1}{x^3}}{x^4 + \frac{1}{x^4}}.$

    So the integral becomes

    $\int \frac{x + \frac{1}{x^3}}{x^4 + \frac{1}{x^4}} \, dx.$

    2. Rewrite the denominator in terms of $x^2$: Notice that

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

    This is a standard algebraic identity: $(a+b)^2 = a^2 + 2ab + b^2$, so with $a = x^2$, $b = 1/x^2$, we get $x^4 + 2 + 1/x^4$, hence $x^4 + 1/x^4 = (x^2 + 1/x^2)^2 - 2$.

    3. Now consider the substitution $t = x^4 - \frac{1}{x^4}$. Differentiate:

    $\frac{dt}{dx} = 4x^3 + \frac{4}{x^5} = 4\left(x^3 + \frac{1}{x^5}\right).$

    That doesn’t match our numerator $x + 1/x^3$ directly. But we can also try $u = x^2 - \frac{1}{x^2}$? Let’s check:

    $\frac{du}{dx} = 2x + \frac{2}{x^3} = 2\left(x + \frac{1}{x^3}\right).$

    That’s exactly twice our numerator! So the substitution $u = x^2 - \frac{1}{x^2}$ is promising.

    4. Express the denominator in terms of $u$: We have $u = x^2 - \frac{1}{x^2}$. Then

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

    So the denominator becomes $u^2 + 2$.

    5. Rewrite the integral: From step 1, the integral is

    $\int \frac{x + \frac{1}{x^3}}{x^4 + \frac{1}{x^4}} \, dx.$

    With $u = x^2 - \frac{1}{x^2}$, we have $du = 2\left(x + \frac{1}{x^3}\right) dx$, so $x + \frac{1}{x^3} \, dx = \frac{du}{2}$.

    And $x^4 + \frac{1}{x^4} = u^2 + 2$. Hence

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

    6. Evaluate the standard integral: …

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12. TG EAPCET 2023Set eng-2023-05-13-FN1 markMCQ
    Q.$\int \frac{1}{16 - 7 \sin^2 x} dx =$ (A) $\frac{1}{12} \tan^{-1} \left( \frac{3 \tan x}{4} \right) + c$ (B) $\frac{1}{3} \sin^{-1} \left( \frac{3 \sin x}{4} \right) + c$ (C) $\frac{1}{12} \log \left( \frac{4 - \sqrt{7} \sin x}{4 + \sqrt{7} \sin x} \right) + c$ (D) $\frac{1}{12} \log \left( \frac{4 + \sqrt{7} \sin x}{4 - \sqrt{7} \sin x} \right) + c$

    ›Reveal solutionSolution

    The integral simplifies by dividing numerator and denominator by $\cos^2 x$, converting it into a standard arctangent form. The correct result is $\frac{1}{12} \tan^{-1}\!\left(\frac{3\tan x}{4}\right)+c$, which corresponds to option (A).

    We are asked to evaluate

    $\int \frac{1}{16 - 7 \sin^2 x} \, dx.$

    The integrand is a rational function of $\sin^2 x$. A classic trick for integrals involving $\sin^2 x$ (or $\cos^2 x$) in the denominator is to rewrite everything in terms of $\tan x$, because $\tan x$ has a simple derivative and lets us turn the integral into a rational function.

    Why this works:

    If we divide numerator and denominator by $\cos^2 x$, we get $\sec^2 x$ in the numerator, which is exactly the derivative of $\tan x$. This substitution $t = \tan x$ transforms the integral into a standard form $\int \frac{dt}{a^2 + t^2}$ or $\int \frac{dt}{a^2 - t^2}$, depending on the sign. Here the denominator becomes $16 - 7\sin^2 x$, and after division by $\cos^2 x$ we get $16\sec^2 x - 7\tan^2 x$, which simplifies nicely.

    Let's work through it step by step.

    1. Rewrite the integrand using $\sin^2 x$ in terms of $\tan x$. Recall $\sin^2 x = \frac{\tan^2 x}{1 + \tan^2 x}$. But a more direct method: multiply numerator and denominator by $\sec^2 x$:

    $\frac{1}{16 - 7\sin^2 x} = \frac{\sec^2 x}{16\sec^2 x - 7\tan^2 x}.$

    Since $\sec^2 x = 1 + \tan^2 x$, the denominator becomes:

    $16(1 + \tan^2 x) - 7\tan^2 x = 16 + 16\tan^2 x - 7\tan^2 x = 16 + 9\tan^2 x.$

    So the integral is:

    $\int \frac{\sec^2 x}{16 + 9\tan^2 x} \, dx.$

    2. Substitute $t = \tan x$. Then $dt = \sec^2 x \, dx$, and the integral becomes:

    $\int \frac{dt}{16 + 9t^2}.$

    3. Factor to match the standard arctangent form. Write $16 + 9t^2 = 9\left( \frac{16}{9} + t^2 \right) = 9\left( \left(\frac{4}{3}\right)^2 + t^2 \right)$. So: $\int \frac{dt}{16 + 9t^2} = \frac{1}{9} \int \frac{dt}{t^2 + \left(\frac{4}{3}\right)^2}.$ …

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