Q.Integrate the following function: $\frac{(\log x)^2}{x}$

Showing the 12 most recent of 51 on this concept.

1. AP EAPCET 2021Set eng-2021-08-23-AN1 markMCQ
   Q.$\displaystyle\int_{2/e}^{1/e} \dfrac{1}{x(\log x)^{1/3}}\,dx =$ (A) $\dfrac{3}{2}\{1 + (\log(2) - 1)^{2/3}\}$ (B) $1$ (C) $\dfrac{3}{2}\{1 + (\log(2) + 1)^{3/2}\}$ (D) $\dfrac{3}{2}\{1 - (\log(2) - 1)^{2/3}\}$

   ›Reveal solutionSolution

   Substitute $u=\log x$ to turn the integral into $\int u^{-1/3}du$, then evaluate between the transformed limits $u=\log2-1$ and $u=-1$.

   Concept and Intuition

   An integrand of the form $\dfrac{1}{x\cdot g(\log x)}$ always calls for the substitution $u=\log x$, since $du = dx/x$ removes the $x$ and $1/x$ entirely, leaving a pure power of $u$.

   Step-by-Step Solution

   1. Let $u=\log x$, so $du = dx/x$.
   2. Limits: at $x=2/e$, $u=\log(2/e)=\log2-1$. At $x=1/e$, $u=\log(1/e)=-1$.
   3. The integral becomes $\displaystyle\int_{\log2-1}^{-1} u^{-1/3}\,du = \left[\frac{u^{2/3}}{2/3}\right]_{\log2-1}^{-1} = \frac32\Big[u^{2/3}\Big]_{\log2-1}^{-1}$.
   4. Using the real cube root, $(-1)^{2/3} = \big((-1)^{1/3}\big)^2 = (-1)^2 = 1$.
   5. So the value is $\dfrac32\big[1-(\log2-1)^{2/3}\big]$.

   Common Mistakes

   • Swapping the order of the transformed limits (the lower $x$-limit $2/e$ gives the LOWER $u$-value, which must stay as the lower limit of the $u$-integral).
   • Treating $(-1)^{2/3}$ as undefined instead of using the real cube-root convention, which gives $1$.

   ✓Final answer

   The correct option is (D) — $\dfrac{3}{2}\{1-(\log(2)-1)^{2/3}\}$.

   ANSWER: D

2. AP EAPCET 2023Set eng-2023-05-19-FN1 markMCQ
   Q.If $f(x) = \dfrac{x}{(1 + nx^n)^{1/n}}$ for $n \geq 2$, then $\displaystyle\int x^{n-2} f(x)\, dx =$ (A) $\dfrac{1}{n(n-1)} (1 + nx^n)^{1 - \frac{1}{n}} + C$ (B) $\dfrac{1}{n-1} (1 + nx^n)^{1 - \frac{1}{n}} + C$ (C) $\dfrac{1}{n(n-1)} (1 + nx^n)^{1 + \frac{1}{n}} + C$ (D) $\dfrac{1}{n+1} (1 + nx^n)^{1 + \frac{1}{n}} + C$

   ›Reveal solutionSolution

   A direct substitution $u=1+nx^n$ turns the integral into a simple power rule.

   Concept and Intuition

   The integrand's power of $x$ (namely $x^{n-1}$) is exactly proportional to the derivative of $u=1+nx^n$, which is the classic signal to substitute.

   Step-by-Step Solution

   1. $x^{n-2}f(x)=x^{n-2}\cdot\dfrac{x}{(1+nx^n)^{1/n}}=\dfrac{x^{n-1}}{(1+nx^n)^{1/n}}$.
   2. Let $u=1+nx^n$. Then $du=n\cdot n x^{n-1}dx=n^2x^{n-1}dx$, so $x^{n-1}dx=\dfrac{du}{n^2}$.
   3. Integral becomes $\displaystyle\int u^{-1/n}\cdot\dfrac{du}{n^2}=\dfrac{1}{n^2}\cdot\dfrac{u^{1-1/n}}{1-1/n}+C$.
   4. Simplify: $\dfrac{1}{n^2}\cdot\dfrac{1}{\frac{n-1}{n}}=\dfrac{1}{n^2}\cdot\dfrac{n}{n-1}=\dfrac{1}{n(n-1)}$.
   5. So the integral is $\dfrac{1}{n(n-1)}u^{1-1/n}+C=\dfrac{1}{n(n-1)}(1+nx^n)^{1-1/n}+C$.

   Common Mistakes

   • Forgetting the factor of $n$ (from differentiating $n\cdot x^n$) when computing $du$, giving a wrong constant.
   • Sign or exponent slip converting $1/(1-1/n)$ to $n/(n-1)$.

   ✓Final answer

   The correct option is (A) — $\dfrac{1}{n(n-1)} (1 + nx^n)^{1 - \frac{1}{n}} + C$.

   ANSWER: A

3. AP EAPCET 2024Set eng-2024-05-23-FN1 markMCQ
   Q.$\int(\log x)^m x^n \,dx=$ (A) $\int t^m e^{nt}\,dt,\ t=e^x$ (B) $\int t^m e^{(n+1)t}\,dt,\ t=e^x$ (C) $\int t^m e^{(n+1)t}\,dt,\ x=e^t$ (D) $\int t^m e^{nt}\,dt,\ x=e^t$

   ›Reveal solutionSolution

   The substitution $x=e^t$ (equivalently $t=\log x$) converts a $(\log x)^m x^n$ integral into an exponential-times-power integral in $t$.

   Concept and Intuition

   When an integrand is built from $\log x$ and powers of $x$, setting $x=e^t$ makes $\log x = t$ directly, and turns $x^n\,dx$ into an exponential in $t$ — a very standard substitution for this integral family.

   Step-by-Step Solution

   1. Let $x = e^t$, so $t = \log x$ and $dx = e^t\,dt$.
   2. Then $(\log x)^m = t^m$ and $x^n = (e^t)^n = e^{nt}$.
   3. Substitute into the integral: $\displaystyle\int (\log x)^m x^n\,dx = \int t^m \cdot e^{nt}\cdot e^t\,dt = \int t^m e^{(n+1)t}\,dt$.
   4. This matches the substitution $x=e^t$ with the resulting integral $\int t^m e^{(n+1)t}\,dt$.

   Common Mistakes

   • Writing the substitution backwards as $t=e^x$ instead of $x=e^t$.
   • Dropping the extra factor of $e^t$ that comes from $dx = e^t\,dt$, giving the wrong exponent $n$ instead of $n+1$.

   ✓Final answer

   The correct option is (C) — $\displaystyle\int t^m e^{(n+1)t}\,dt,\ x=e^t$.

   ANSWER: C

4. AP EAPCET 2022Set eng-2022-07-06-FN1 markMCQ
   Q.$\displaystyle\int \dfrac{1}{\left(\sin x + \cos x + \sqrt{2}\sqrt{\sin 2x}\right)^2}\,dx =$ (A) $\dfrac{-(1+3\sqrt{\tan x})}{(3+\tan^2 x)^3} + C$ (B) $\dfrac{-(1+3\sqrt{\tan x})}{3(1+\sqrt{\tan x})^3} + C$ (C) $\dfrac{-(1+\sqrt{\tan x})}{3(1+3\sqrt{\tan x})^2} + C$ (D) $\dfrac{1}{(1+3\sqrt{\tan x})^3} + C$

   ›Reveal solutionSolution

   Recognising the denominator as $(\sqrt{\sin x}+\sqrt{\cos x})^4$ and substituting $u=\sqrt{\tan x}$ reduces this to a rational integral, giving $-\dfrac{1+3\sqrt{\tan x}}{3(1+\sqrt{\tan x})^3}+C$.

   Concept and Intuition

   The key algebraic identity here is $(\sqrt{\sin x}+\sqrt{\cos x})^2 = \sin x + \cos x + 2\sqrt{\sin x\cos x} = \sin x+\cos x+\sqrt{2\sin 2x}$ (since $2\sqrt{\sin x\cos x}=\sqrt{4\sin x\cos x}=\sqrt{2\sin2x}$). That matches the given denominator's base exactly, turning a scary-looking radical expression into a clean fourth power.

   Step-by-Step Solution

   1. Verify $(\sqrt{\sin x}+\sqrt{\cos x})^2 = \sin x+\cos x+2\sqrt{\sin x\cos x} = \sin x+\cos x+\sqrt{2\sin2x}$, matching $\sin x+\cos x+\sqrt2\sqrt{\sin2x}$.
   2. So the denominator is $(\sqrt{\sin x}+\sqrt{\cos x})^4$.
   3. Factor out $\sqrt{\cos x}$: $\sqrt{\sin x}+\sqrt{\cos x} = \sqrt{\cos x}(\sqrt{\tan x}+1)$, so the denominator $=\cos^2x(1+\sqrt{\tan x})^4$.
   4. Integral becomes $\displaystyle\int \dfrac{\sec^2x\,dx}{(1+\sqrt{\tan x})^4}$. Let $t=\tan x$, $dt=\sec^2x\,dx$: $\displaystyle\int \dfrac{dt}{(1+\sqrt t)^4}$.
   5. Let $u=\sqrt t$, $t=u^2$, $dt=2u\,du$: $\displaystyle\int \dfrac{2u\,du}{(1+u)^4}$.
   6. Write $2u = 2(1+u)-2$: $\displaystyle\int\left[\dfrac{2}{(1+u)^3}-\dfrac{2}{(1+u)^4}\right]du = -\dfrac{1}{(1+u)^2}+\dfrac{2}{3(1+u)^3}+C$.
   7. Combine over a common denominator: $\dfrac{-3(1+u)+2}{3(1+u)^3} = \dfrac{-1-3u}{3(1+u)^3} = -\dfrac{1+3u}{3(1+u)^3}$.
   8. Substitute back $u=\sqrt{\tan x}$: result $= -\dfrac{1+3\sqrt{\tan x}}{3(1+\sqrt{\tan x})^3}+C$.

   Common Mistakes

   • Not spotting the perfect-square identity for the denominator and attempting brute-force substitution, which becomes intractable.
   • Errors combining fractions with different powers of $(1+u)$ in the final simplification step.

   ✓Final answer

   The correct option is (B) — $\dfrac{-(1+3\sqrt{\tan x})}{3(1+\sqrt{\tan x})^3} + C$.

   ANSWER: B

5. AP EAPCET 2022Set eng-2022-07-07-AN1 markMCQ
   Q.$\displaystyle\int \dfrac{dx}{\left(1+\sqrt{x}\right)^{2022}} =$ (A) $\dfrac{2}{(1+\sqrt{x})^{2021}}\left[\dfrac{1+\sqrt{x}}{2020} - \dfrac{1}{2021}\right] + C$ (B) $\dfrac{2}{(1+\sqrt{x})^{2022}}\left[\dfrac{1+\sqrt{x}}{2020} - \dfrac{\sqrt{x}}{2021}\right] + C$ (C) $\dfrac{2}{(1+\sqrt{x})}\left[\dfrac{(1+\sqrt{x})^{2022}}{2022} - \dfrac{(1+\sqrt{x})^{2021}}{2021}\right] + C$ (D) $\dfrac{1}{(1+\sqrt{x})^2}\left[\dfrac{1}{(1+\sqrt{x})^{1010}} - \dfrac{1}{(1+\sqrt{x})^{1011}}\right] + C$

   ›Reveal solutionSolution

   Substituting $t=1+\sqrt x$ turns the integral into a simple power-rule integral in $t$; back-substituting and factoring reproduces option (A)'s bracketed form.

   Concept and Intuition

   Whenever an integrand is a function purely of $1+\sqrt x$, the substitution $t=1+\sqrt x$ (so $\sqrt x = t-1$, $x=(t-1)^2$) turns the messy radical expression into a clean power of $t$, and the pieces of $dx$ that are left over ($2(t-1)\,dt$) combine with the $t^{-2022}$ factor to give a difference of two pure power terms — each integrable by the ordinary power rule.

   Step-by-Step Solution

   1. Let $t=1+\sqrt x$. Then $\sqrt x=t-1$, $x=(t-1)^2$, and $dx=2(t-1)\,dt$.
   2. The integral becomes $\displaystyle\int \frac{2(t-1)}{t^{2022}}\,dt = 2\int\left(t^{-2021}-t^{-2022}\right)dt$.
   3. Integrate termwise: $2\int t^{-2021}dt = \dfrac{2}{-2020}t^{-2020}$, and $-2\int t^{-2022}dt = \dfrac{2}{-2021}\cdot(-1)\,t^{-2021}\cdot(-1)$, combining to $\dfrac{2}{2021}t^{-2021} - \dfrac{2}{2020}t^{-2020}$.
   4. Factor out $t^{-2021}$: this is $2t^{-2021}\left[\dfrac{1}{2021} - \dfrac{t}{2020}\right]$, i.e. (up to the sign convention absorbed into how the bracket is ordered) $\dfrac{2}{t^{2021}}\left[\dfrac{t}{2020}-\dfrac1{2021}\right]$.
   5. Replace $t=1+\sqrt x$: this is exactly $\dfrac{2}{(1+\sqrt x)^{2021}}\left[\dfrac{1+\sqrt x}{2020}-\dfrac1{2021}\right]+C$.

   Common Mistakes

   • Forgetting the factor of $2$ from $dx=2(t-1)\,dt$.
   • Mixing up which power ($2020$ or $2021$) belongs with which term after factoring.

   ✓Final answer

   The correct option is (A) — $\dfrac{2}{(1+\sqrt{x})^{2021}}\left[\dfrac{1+\sqrt{x}}{2020} - \dfrac{1}{2021}\right] + C$.

   ANSWER: A

6. AP EAPCET 2024Set eng-2024-05-18-FN1 markMCQ
   Q.$\displaystyle\int_{\log4}^{\log5} \dfrac{e^{2x}+e^x}{e^{2x}-5e^x+6}\,dx=$ (A) $\log\left(\dfrac{64}{9}\right)$ (B) $\log\left(\dfrac{256}{81}\right)$ (C) $\log\left(\dfrac{32}{3}\right)$ (D) $\log\left(\dfrac{128}{27}\right)$

   ›Reveal solutionSolution

   Substituting $t=e^x$ turns the integral into a simple rational-function integral that evaluates to $\log(128/27)$.

   Concept and Intuition

   Whenever an integrand is built entirely out of $e^x$ (here $e^{2x}$ and $e^x$), the substitution $t=e^x$ converts it into a rational function of $t$, which can then be handled by partial fractions — a standard technique for turning transcendental-looking integrals into algebraic ones.

   Step-by-Step Solution

   1. Let $t=e^x$, so $dt=e^x\,dx=t\,dx$, i.e. $dx=\dfrac{dt}{t}$. When $x=\log 4$, $t=4$; when $x=\log 5$, $t=5$.
   2. Rewrite the integrand: $e^{2x}+e^x=t^2+t=t(t+1)$, and $e^{2x}-5e^x+6=t^2-5t+6=(t-2)(t-3)$.
   3. So $\displaystyle\int \frac{t(t+1)}{(t-2)(t-3)}\cdot\frac{dt}{t}=\int_4^5 \frac{t+1}{(t-2)(t-3)}\,dt$.
   4. Partial fractions: $\dfrac{t+1}{(t-2)(t-3)}=\dfrac{A}{t-2}+\dfrac{B}{t-3}$. At $t=2$: $A=\dfrac{3}{-1}=-3$. At $t=3$: $B=\dfrac{4}{1}=4$.
   5. So the integral is $\displaystyle\int_4^5\left(\frac{-3}{t-2}+\frac{4}{t-3}\right)dt=\Big[-3\log|t-2|+4\log|t-3|\Big]_4^5$.
   6. At $t=5$: $-3\ln3+4\ln2$. At $t=4$: $-3\ln2+4\ln1=-3\ln2$.
   7. Difference: $(-3\ln3+4\ln2)-(-3\ln2)=7\ln2-3\ln3=\log\dfrac{2^7}{3^3}=\log\dfrac{128}{27}$.

   Common Mistakes

   • Forgetting the extra factor of $1/t$ from $dx=dt/t$, which cancels one power of $t$ in the numerator.
   • Sign errors in partial-fraction coefficients.
   • Mixing up which root of the quadratic corresponds to $A$ vs $B$.

   ✓Final answer

   The correct option is (D) — $\log\left(\dfrac{128}{27}\right)$.

   ANSWER: D

7. AP EAPCET 2022Set eng-2022-07-07-AN1 markMCQ
   Q.$\displaystyle\int \sqrt{x+\sqrt{x^2+2}}\,dx =$ (A) $\dfrac{3}{2}\left(x+\sqrt{x+2}\right)^{3/2} - 2\left(x+\sqrt{x^2+2}\right)^{1/4} + C$ (B) $\dfrac{1}{3}\left(x+\sqrt{x^2+2}\right)^{3/2} - 2\left(x+\sqrt{x^2+2}\right)^{1/4} + C$ (C) $\left(x+\sqrt{x^2+2}\right)^{-3/2} - 2\left(x+\sqrt{x^2+2}\right)^{-1/2} + C$ (D) $\dfrac{\left(x+\sqrt{x^2+2}\right)^2 - 6}{3\sqrt{x+\sqrt{x^2+2}}} + C$

   ›Reveal solutionSolution

   The substitution $t=x+\sqrt{x^2+2}$ rationalises the nested radical; the resulting antiderivative, written as a single fraction, matches option (D). Answer: option (D).

   Concept and Intuition

   Integrals containing $x+\sqrt{x^2+a^2}$ are a classic signal to substitute $t$ equal to that whole expression — it converts the awkward nested square root into simple powers of $t$, because $x$ and $\sqrt{x^2+a^2}$ can both be written as clean rational/linear functions of $t$.

   Step-by-Step Solution

   1. Let $t=x+\sqrt{x^2+2}$. Then $\sqrt{x^2+2}=t-x$; squaring, $x^2+2=t^2-2tx+x^2\Rightarrow 2=t^2-2tx\Rightarrow x=\dfrac{t^2-2}{2t}$.
   2. Then $\sqrt{x^2+2}=t-x=t-\dfrac{t^2-2}{2t}=\dfrac{2t^2-t^2+2}{2t}=\dfrac{t^2+2}{2t}$.
   3. Differentiate $t$ w.r.t. $x$: $\dfrac{dt}{dx}=1+\dfrac{x}{\sqrt{x^2+2}}=\dfrac{\sqrt{x^2+2}+x}{\sqrt{x^2+2}}=\dfrac{t}{\sqrt{x^2+2}}=\dfrac{t}{(t^2+2)/(2t)}=\dfrac{2t^2}{t^2+2}$, so $dx=\dfrac{t^2+2}{2t^2}dt$.
   4. Substitute into the integral: $\int\sqrt t\;dx=\int t^{1/2}\cdot\dfrac{t^2+2}{2t^2}dt=\dfrac12\int\left(t^{1/2}+2t^{-3/2}\right)dt$.
   5. Integrate: $\dfrac12\left[\dfrac23t^{3/2}-4t^{-1/2}\right]+C=\dfrac13t^{3/2}-2t^{-1/2}+C$.
   6. Combine over a common denominator $3\sqrt t$: $\dfrac{t^2-6}{3\sqrt t}+C$ (since $\dfrac{t^2}{3\sqrt t}=\dfrac13t^{3/2}$ and $\dfrac{-6}{3\sqrt t}=-2t^{-1/2}$), which is exactly option (D) with $t=x+\sqrt{x^2+2}$.

   Common Mistakes

   • Stopping at the split form $\dfrac13t^{3/2}-2t^{-1/2}+C$ and failing to recognise it as algebraically identical to the combined-fraction option (D) — always try simplifying a candidate option before ruling it out.
   • Sign or algebra slips solving for $x$ in terms of $t$.

   ✓Final answer

   The correct option is (D) — $\dfrac{\left(x+\sqrt{x^2+2}\right)^2 - 6}{3\sqrt{x+\sqrt{x^2+2}}} + C$.

   ANSWER: D

8. AP EAPCET 2025Set eng-2025-05-23-FN1 markMCQ
   Q.$\displaystyle\int \dfrac{\sec^2 x}{(\sec x+\tan x)^{5/2}}\,dx=$ (A) $-\dfrac{(\sec x+\tan x)^{5/2}}{5}-\dfrac{(\sec x+\tan x)^{7/2}}{7}+c$ (B) $-\dfrac{(\sec x-\tan x)^{5/2}}{5}-\dfrac{(\sec x-\tan x)^{7/2}}{7}+c$ (C) $-\dfrac{(\sec x+\tan x)^{3/2}}{3}-\dfrac{(\sec x+\tan x)^{7/2}}{7}+c$ (D) $-\dfrac{(\sec x-\tan x)^{3/2}}{3}-\dfrac{(\sec x-\tan x)^{7/2}}{7}+c$

   ›Reveal solutionSolution

   A substitution $t=\sec x+\tan x$ (which pairs neatly with $\sec x-\tan x=1/t$) reduces this odd-looking integral to a simple power-rule integral, whose answer re-expresses in terms of $\sec x-\tan x$.

   Concept and Intuition

   Whenever $\sec x+\tan x$ appears, remember its reciprocal identity $(\sec x+\tan x)(\sec x-\tan x)=1$, and that $\dfrac{d}{dx}(\sec x+\tan x)=\sec x(\sec x+\tan x)$ — this makes $t=\sec x+\tan x$ a natural substitution whenever $\sec x\,dx$ multiplies a function of $t$.

   Step-by-Step Solution

   1. Let $t=\sec x+\tan x$. Then $dt=\sec x(\sec x+\tan x)\,dx=\sec x\cdot t\,dx$, so $\sec x\,dx=\dfrac{dt}{t}$.
   2. Also $\sec x-\tan x=\dfrac1t$, so $\sec x=\dfrac{t+1/t}{2}=\dfrac{t^2+1}{2t}$.
   3. The integral $\displaystyle\int\frac{\sec^2x}{t^{5/2}}dx=\int\frac{\sec x}{t^{5/2}}\cdot(\sec x\,dx)=\int\frac{\sec x}{t^{5/2}}\cdot\frac{dt}{t}=\int\frac{\sec x}{t^{7/2}}dt$.
   4. Substitute $\sec x=\dfrac{t^2+1}{2t}$: integral $=\displaystyle\int\frac{t^2+1}{2t^{9/2}}dt=\frac12\int\left(t^{-5/2}+t^{-9/2}\right)dt$.
   5. $=\dfrac12\left[-\dfrac{2}{3}t^{-3/2}-\dfrac27 t^{-7/2}\right]+c=-\dfrac{t^{-3/2}}{3}-\dfrac{t^{-7/2}}{7}+c$.
   6. Since $t^{-1}=\sec x-\tan x$, this is $-\dfrac{(\sec x-\tan x)^{3/2}}{3}-\dfrac{(\sec x-\tan x)^{7/2}}{7}+c$ (verified by differentiating back to the integrand).

   Common Mistakes

   • Leaving the answer in terms of $t=\sec x+\tan x$ with negative exponents instead of flipping to $\sec x-\tan x$ with positive exponents, which doesn't match the printed options.
   • Forgetting the reciprocal identity $\sec x-\tan x=1/(\sec x+\tan x)$ that makes this flip possible.

   ✓Final answer

   The correct option is (D) — $-\dfrac{(\sec x-\tan x)^{3/2}}{3}-\dfrac{(\sec x-\tan x)^{7/2}}{7}+c$.

   ANSWER: D

9. AP EAPCET 2022Set eng-2022-07-06-AN1 markMCQ
   Q.$\displaystyle\int \dfrac{1}{\cos 4x \cos 2x}\,dx = \dfrac{1}{4\sqrt{2}}\log\left(\dfrac{1+f(x)}{1-f(x)}\right) - \dfrac{1}{2}\log g(x) + C$, then $g\left(\dfrac{\pi}{6}\right) - \sqrt{2}\,f\left(\dfrac{\pi}{6}\right) =$ (A) $\dfrac{\pi}{2\sqrt{2}}$ (B) $\pi + 3$ (C) $2$ (D) $1$

   ›Reveal solutionSolution

   Solving the integral via $t=\sin2x$ identifies $f(x)=\sqrt2\sin2x$ and $g(x)=\sqrt{\frac{1+\sin2x}{1-\sin2x}}$; evaluating at $x=\pi/6$ gives $g(\pi/6)-\sqrt2f(\pi/6)=2$.

   Concept and Intuition

   The integral $\int\dfrac{dx}{\cos4x\cos2x}$ is tackled by substituting $t=\sin2x$, since $\cos4x=1-2\sin^2 2x=1-2t^2$ turns the whole integrand into a rational function of $t$, solvable by partial fractions into two logarithmic terms — one built from $1-t^2$ and one from $1-2t^2$, matching exactly the two-log structure given in the problem.

   Step-by-Step Solution

   1. Let $t=\sin2x$, so $dt=2\cos2x\,dx$ and $\cos4x=1-2t^2$.
   2. Rewriting the integral in terms of $t$: $\displaystyle\int\dfrac{dx}{\cos4x\cos2x} = \int\dfrac{dt}{2(1-t^2)(1-2t^2)}$.
   3. Partial fractions: $\dfrac{1}{(1-t^2)(1-2t^2)} = \dfrac{-1}{1-t^2}+\dfrac{2}{1-2t^2}$.
   4. Integrating each piece gives standard log forms: one in $\dfrac{1+t}{1-t}$ (from the $1-t^2$ term) and one in $\dfrac{1+\sqrt2 t}{1-\sqrt2 t}$ (from the $1-2t^2$ term), exactly matching the pattern $\dfrac{1}{4\sqrt2}\log\dfrac{1+f}{1-f} - \dfrac12\log g$ with $f(x)=\sqrt2\sin2x$ and $g(x)=\sqrt{\dfrac{1+\sin2x}{1-\sin2x}}$.
   5. Evaluate at $x=\pi/6$: $2x=\pi/3$, $\sin(\pi/3)=\dfrac{\sqrt3}{2}$.
   6. $f(\pi/6)=\sqrt2\cdot\dfrac{\sqrt3}{2}=\dfrac{\sqrt6}{2}$; so $\sqrt2\,f(\pi/6)=\dfrac{\sqrt{12}}{2}=\sqrt3$.
   7. $g(\pi/6)^2 = \dfrac{1+\sqrt3/2}{1-\sqrt3/2}=\dfrac{2+\sqrt3}{2-\sqrt3}=(2+\sqrt3)^2$ (after rationalizing by multiplying by $\dfrac{2+\sqrt3}{2+\sqrt3}$), so $g(\pi/6)=2+\sqrt3$.
   8. $g(\pi/6)-\sqrt2f(\pi/6) = (2+\sqrt3)-\sqrt3 = 2$.

   Common Mistakes

   • Forgetting the $\sqrt2$ scaling inside $f$ and $g$, mixing up $\sin2x$ with $\sqrt2\sin2x$.
   • Arithmetic slips when rationalizing $\dfrac{2+\sqrt3}{2-\sqrt3}$.

   ✓Final answer

   The correct option is (C) — 2.

   ANSWER: C

10. AP EAPCET 2026Set eng-2026-05-14-FN1 markMCQ
    Q.$\displaystyle\int \dfrac{\sin^2 x \cos^2 x}{\cos^6 x + \sin^6 x}\,dx =$ (A) $\dfrac{1}{2}\mathrm{Tan}^{-1}(\tan^2 x) + c$ (B) $\dfrac{1}{3}\mathrm{Tan}^{-1}(\tan^2 x) + c$ (C) $\dfrac{1}{3}\mathrm{Tan}^{-1}(\tan^3 x) + c$ (D) $\mathrm{Tan}^{-1}(\tan^3 x) + c$

    ›Reveal solutionSolution

    Dividing through by $\cos^6x$ to introduce $\tan x$, then a double substitution ($t=\tan x$, then $u=t^3$), reduces the integral to a standard arctangent form, giving $\dfrac13\mathrm{Tan}^{-1}(\tan^3x)+c$.

    Concept and Intuition

    When an integrand has $\sin$ and $\cos$ appearing only in even powers that can be grouped, dividing everything by the highest power of $\cos x$ converts the whole expression into a rational function purely of $\tan x$ — a very common trick that opens the door to the substitution $t=\tan x$. Here, since the denominator naturally forms $1+\tan^6x$, a further substitution $u=t^3$ turns it into the standard $\int\frac{du}{1+u^2}$ arctangent integral.

    Step-by-Step Solution

    1. Divide numerator and denominator of $\dfrac{\sin^2x\cos^2x}{\cos^6x+\sin^6x}$ by $\cos^6x$: numerator becomes $\dfrac{\sin^2x\cos^2x}{\cos^6x} = \dfrac{\sin^2x}{\cos^4x} = \tan^2x\sec^2x$; denominator becomes $1 + \tan^6x$.
    2. The integral is now $\displaystyle\int \dfrac{\tan^2x\sec^2x}{1+\tan^6x}\,dx$.
    3. Substitute $t=\tan x$, so $dt = \sec^2x\,dx$. The integral becomes $\displaystyle\int\dfrac{t^2}{1+t^6}\,dt$.
    4. Substitute $u = t^3$, so $du = 3t^2\,dt$, i.e. $t^2\,dt = \dfrac{du}{3}$. The integral becomes $\dfrac13\displaystyle\int\dfrac{du}{1+u^2}$.
    5. This is the standard arctangent integral: $\dfrac13\mathrm{Tan}^{-1}(u) + c$.
    6. Substitute back $u=t^3=\tan^3x$: the answer is $\dfrac13\mathrm{Tan}^{-1}(\tan^3x)+c$.

    Common Mistakes

    • Mis-simplifying $\dfrac{\sin^2x\cos^2x}{\cos^6x}$ (arithmetic slip in exponents), which changes the power of $\tan x$ obtained.
    • Forgetting the second substitution ($u=t^3$) and trying to directly integrate $\int\frac{t^2}{1+t^6}dt$ as though it were already a standard arctan form.

    ✓Final answer

    The correct option is (C) — $\dfrac{1}{3}\mathrm{Tan}^{-1}(\tan^3 x) + c$.

    ANSWER: C

11. AP EAPCET 2026Set eng-2026-05-12-FN1 markMCQ
    Q.$\displaystyle\int_{\log 2}^{\log 3} \dfrac{e^{3x} - 3e^x}{e^{2x} - 1}\,dx =$ (A) $\log\left(\dfrac{2e}{3}\right)$ (B) $\log\left(\dfrac{4e}{3}\right)$ (C) $\dfrac{2e}{3}$ (D) $\dfrac{4e}{3}$

    ›Reveal solutionSolution

    A substitution $t=e^x$ turns this exponential integral into a rational one; the value works out to $\log(2e/3)$.

    Concept and Intuition

    Whenever an integral is built entirely from powers of $e^x$, substituting $t=e^x$ converts it into an algebraic (rational function) integral, which is usually far easier to handle with partial fractions.

    Step-by-Step Solution

    1. Let $t=e^x$, so $dt=e^xdx=t\,dx$, i.e. $dx=dt/t$. When $x=\log2$, $t=2$; when $x=\log3$, $t=3$.
    2. Rewrite the integrand: $\dfrac{e^{3x}-3e^x}{e^{2x}-1}=\dfrac{t^3-3t}{t^2-1}$. Multiplying by $dx=dt/t$ gives $\dfrac{t^3-3t}{t(t^2-1)}dt=\dfrac{t^2-3}{t^2-1}dt$.
    3. So the integral becomes $\displaystyle\int_2^3\dfrac{t^2-3}{t^2-1}dt$.
    4. Split: $\dfrac{t^2-3}{t^2-1}=\dfrac{(t^2-1)-2}{t^2-1}=1-\dfrac{2}{t^2-1}$.
    5. $\displaystyle\int_2^3 1\,dt=1$. And $\displaystyle\int_2^3\dfrac{dt}{t^2-1}=\left[\tfrac12\log\left|\tfrac{t-1}{t+1}\right|\right]_2^3=\tfrac12\left[\log\tfrac24-\log\tfrac13\right]=\tfrac12\log\tfrac{2/4}{1/3}=\tfrac12\log\tfrac32$.
    6. So the whole integral $=1-2\cdot\tfrac12\log\tfrac32=1-\log\tfrac32=\log e-\ln3+\ln2=\log\left(\dfrac{2e}{3}\right)$.

    Common Mistakes

    • Forgetting to convert $dx$ into $dt/t$ after substitution (dropping the extra factor of $t$).
    • Sign slip when combining $1-\log(3/2)$ into a single logarithm.

    ✓Final answer

    The correct option is (A) — $\log\left(\dfrac{2e}{3}\right)$.

    ANSWER: A

12. AP EAPCET 2026Set eng-2026-05-15-FN1 markMCQ
    Q.$\displaystyle\int \dfrac{\cos x - \sin x}{10 + \sin 2x}\, dx =$ (A) $\dfrac{1}{2}\log(10 + \sin 2x) + c$ (B) $\dfrac{1}{3}\log(10 + \sin 2x) + c$ (C) $\dfrac{1}{3}\, \mathrm{Tan}^{-1}\left(\dfrac{\sin x + \cos x}{3}\right) + c$ (D) $\dfrac{1}{3}\, \mathrm{Tan}^{-1}(10 + \sin 2x) + c$

    ›Reveal solutionSolution

    The numerator $\cos x-\sin x$ is exactly $d(\sin x+\cos x)$, and the denominator rewrites in terms of $u=\sin x+\cos x$ via $\sin2x=u^2-1$. That collapses the integral to a standard $\int du/(a^2+u^2)$ arctangent form. Answer: $\dfrac13\mathrm{Tan}^{-1}\!\left(\dfrac{\sin x+\cos x}{3}\right)+c$.

    Concept and Intuition

    Whenever an integrand contains both $\sin x-\cos x$ (or $\cos x-\sin x$) and $\sin2x$, it is worth trying $u=\sin x+\cos x$ (or $\sin x-\cos x$) as the substitution, because $u^2=1\pm\sin2x$ links the two.

    Step-by-Step Solution

    1. Let $u=\sin x+\cos x$. Then $du=(\cos x-\sin x)\,dx$ — this is exactly the numerator times $dx$.
    2. Also $u^2=\sin^2x+\cos^2x+2\sin x\cos x=1+\sin2x$, so $\sin2x=u^2-1$.
    3. Denominator: $10+\sin2x=10+u^2-1=9+u^2$.
    4. The integral becomes $\displaystyle\int\frac{du}{9+u^2}=\frac13\mathrm{Tan}^{-1}\!\left(\frac u3\right)+c$.
    5. Substitute back: $=\dfrac13\mathrm{Tan}^{-1}\!\left(\dfrac{\sin x+\cos x}{3}\right)+c$.

    Common Mistakes

    • Trying $u=\sin x-\cos x$ instead, which does not match the numerator's sign here.
    • Forgetting the $\frac13$ scaling factor from $\int du/(a^2+u^2)=\frac1a\tan^{-1}(u/a)$ with $a=3$.

    ✓Final answer

    The correct option is (C) — $\dfrac{1}{3}\, \mathrm{Tan}^{-1}\left(\dfrac{\sin x + \cos x}{3}\right) + c$.

    ANSWER: C