Q.Integrate the following function: $\frac{1}{x+x \log x}$

Showing the 12 most recent of 51 on this concept.

1. 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

2. 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

3. 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

4. AP EAPCET 2023Set eng-2023-05-15-AN1 markMCQ
   Q.$\displaystyle\int \dfrac{dx}{4+5\cos x} =$ (A) $-\dfrac{1}{3}\log\left|\dfrac{3+\tan\frac{x}{2}}{3-\tan\frac{x}{2}}\right|+C$ (B) $\dfrac{1}{3}\log\left|\dfrac{3+\tan\frac{x}{2}}{3-\tan\frac{x}{2}}\right|+C$ (C) $-\dfrac{1}{9}\log\left|\dfrac{3-\tan\frac{x}{2}}{3+\tan\frac{x}{2}}\right|+C$ (D) $\dfrac{1}{9}\log\left|\dfrac{3-\tan\frac{x}{2}}{3+\tan\frac{x}{2}}\right|+C$

   ›Reveal solutionSolution

   The Weierstrass substitution $t=\tan(x/2)$ turns this into a standard $\int dt/(a^2-t^2)$ integral, giving option (B).

   Concept and Intuition

   For $\int\dfrac{dx}{a+b\cos x}$ the substitution $t=\tan(x/2)$ (so $\cos x=\tfrac{1-t^2}{1+t^2}$, $dx=\tfrac{2\,dt}{1+t^2}$) always converts the integral into a rational function of $t$ alone.

   Step-by-Step Solution

   1. $\cos x=\dfrac{1-t^2}{1+t^2}$, $dx=\dfrac{2\,dt}{1+t^2}$, with $t=\tan(x/2)$.
   2. $4+5\cos x=4+5\cdot\dfrac{1-t^2}{1+t^2}=\dfrac{4(1+t^2)+5(1-t^2)}{1+t^2}=\dfrac{9-t^2}{1+t^2}$.
   3. Integral becomes $\int\dfrac{2\,dt/(1+t^2)}{(9-t^2)/(1+t^2)}=\int\dfrac{2\,dt}{9-t^2}$.
   4. Using $\int\dfrac{dt}{a^2-t^2}=\dfrac{1}{2a}\log\left|\dfrac{a+t}{a-t}\right|+C$ with $a=3$: $\int\dfrac{dt}{9-t^2}=\dfrac16\log\left|\dfrac{3+t}{3-t}\right|+C$.
   5. Multiply by 2: $\dfrac13\log\left|\dfrac{3+t}{3-t}\right|+C$, i.e. $\dfrac13\log\left|\dfrac{3+\tan(x/2)}{3-\tan(x/2)}\right|+C$.

   Common Mistakes

   • Sign slip giving $-\dfrac13\log|\ldots|$ instead of $+\dfrac13$.
   • Using $\dfrac1{2a}$ instead of doubling it correctly after the substitution factor of 2.

   ✓Final answer

   The correct option is (B) — $\dfrac{1}{3}\log\left|\dfrac{3+\tan\frac{x}{2}}{3-\tan\frac{x}{2}}\right|+C$.

   ANSWER: B

5. 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

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 2024Set eng-2024-05-18-FN1 markMCQ
   Q.$\displaystyle\int(\tan^7x+\tan x)\,dx=$ (A) $\dfrac{\tan^2 x}{12}(2\tan^4 x - 3\tan^2 x+6)+c$ (B) $\dfrac{\tan^2 x}{6}-\dfrac{\tan^5 x}{4}+\dfrac{\tan^4 x}{2}+c$ (C) $\dfrac{\tan^2 x}{6}(\tan^4 x+3\tan^2 x+4)+c$ (D) $\dfrac{\tan x}{12}(\tan^4 x-3\tan^2 x+6)+c$

   ›Reveal solutionSolution

   Factoring $\tan^7x+\tan x$ using the sum-of-like-terms identity $t^7+t=t(t^6+1)=t(t^2+1)(t^4-t^2+1)$ exposes the $\sec^2x$ needed for a clean $t=\tan x$ substitution. Answer: $\dfrac{\tan^2x}{12}(2\tan^4x-3\tan^2x+6)+c$.

   Concept and Intuition

   Whenever an integrand is built purely from powers of $\tan x$ together with an explicit or hidden $\sec^2x$, substituting $t=\tan x$ turns it into a polynomial integral — the key is recognizing that $t^6+1=(t^2+1)(t^4-t^2+1)$ supplies exactly the $\sec^2x=1+\tan^2x$ factor needed.

   Step-by-Step Solution

   1. $\tan^7x+\tan x=\tan x(\tan^6x+1)=\tan x(\tan^2x+1)(\tan^4x-\tan^2x+1)=\tan x\sec^2x(\tan^4x-\tan^2x+1)$.
   2. Let $t=\tan x$, $dt=\sec^2x\,dx$. Integral becomes $\displaystyle\int t(t^4-t^2+1)dt=\int(t^5-t^3+t)dt$.
   3. $=\dfrac{t^6}{6}-\dfrac{t^4}{4}+\dfrac{t^2}{2}+c=\dfrac{\tan^6x}{6}-\dfrac{\tan^4x}{4}+\dfrac{\tan^2x}{2}+c$.
   4. Factor out $\dfrac{\tan^2x}{12}$: $\dfrac{\tan^2x}{12}\left(2\tan^4x-3\tan^2x+6\right)+c$ (check: $\dfrac{2}{12}=\dfrac16$, $\dfrac{3}{12}=\dfrac14$, $\dfrac{6}{12}=\dfrac12$ — all match).

   Common Mistakes

   • Missing the factorization $t^6+1=(t^2+1)(t^4-t^2+1)$ and instead trying to expand $\tan^7x$ directly, which is much harder.
   • Sign error in the middle term when factoring $t^4-t^2+1$ vs $t^4+t^2+1$.

   ✓Final answer

   The correct option is (A) — $\dfrac{\tan^2 x}{12}(2\tan^4 x - 3\tan^2 x+6)+c$.

   ANSWER: A

8. AP EAPCET 2026Set eng-2026-05-14-AN1 markMCQ
   Q.If $\displaystyle\int \dfrac{1}{\sin^3 x+\cos^3 x}\,dx = A\log\left|\dfrac{\sqrt{2}+t}{\sqrt{2}-t}\right|+B\,\mathrm{Tan}^{-1}(t)+c$, then $\left(\dfrac{B}{A},t\right)=$ (A) $(2\sqrt{2},\sin x+\cos x)$ (B) $\left(\dfrac{\sqrt{2}}{9},\sin x+\cos x\right)$ (C) $\left(\dfrac{\sqrt{2}}{9},\sin x-\cos x\right)$ (D) $(2\sqrt{2},\sin x-\cos x)$

   ›Reveal solutionSolution

   Factoring the sum of cubes and substituting $t=\sin x-\cos x$ turns the trigonometric integral into a clean rational-function integral in $t$, giving $B/A=2\sqrt2$ with $t=\sin x-\cos x$.

   Concept and Intuition

   $\sin^3x+\cos^3x$ factors as a sum of cubes: $(\sin x+\cos x)(1-\sin x\cos x)$. Both $\sin x+\cos x$ and $1-\sin x\cos x$ can be written purely in terms of $u=\sin x-\cos x$, because $(\sin x+\cos x)^2+(\sin x-\cos x)^2=2$ and $1-\sin x\cos x=\tfrac{1+(\sin x-\cos x)^2}{2}$. Crucially, $\dfrac{du}{dx}=\cos x+\sin x$, which is exactly the factor left over after using the second identity — so the whole integral collapses into a rational function of $u$ alone.

   Step-by-Step Solution

   1. Factor: $\sin^3x+\cos^3x=(\sin x+\cos x)(1-\sin x\cos x)$.
   2. Let $u=\sin x-\cos x$. Then $u^2=1-2\sin x\cos x$, so $1-\sin x\cos x=\dfrac{1+u^2}{2}$.
   3. Also $(\sin x+\cos x)^2=1+2\sin x\cos x=2-u^2$, so $\sin x+\cos x=\sqrt{2-u^2}$ (taking the appropriate branch), and $\dfrac{du}{dx}=\cos x+\sin x=\sqrt{2-u^2}$.
   4. So $\sin^3x+\cos^3x=\sqrt{2-u^2}\cdot\dfrac{1+u^2}{2}$, and

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

   5. Partial fractions (by symmetry, only even terms survive): $\dfrac{2}{(2-u^2)(1+u^2)}=\dfrac{2/3}{2-u^2}+\dfrac{2/3}{1+u^2}$.
   6. Integrate: $\displaystyle\int\frac{du}{2-u^2}=\frac{1}{2\sqrt2}\log\left|\frac{\sqrt2+u}{\sqrt2-u}\right|$ and $\int\frac{du}{1+u^2}=\mathrm{Tan}^{-1}u$.
   7. So $I=\dfrac{1}{3\sqrt2}\log\left|\dfrac{\sqrt2+u}{\sqrt2-u}\right|+\dfrac{2}{3}\mathrm{Tan}^{-1}u+c$, matching the given form with $A=\dfrac{1}{3\sqrt2}$, $B=\dfrac{2}{3}$, $t=u=\sin x-\cos x$.
   8. Hence $\dfrac{B}{A}=\dfrac{2/3}{1/(3\sqrt2)}=2\sqrt2$.

   Common Mistakes

   • Trying $t=\sin x+\cos x$ directly — it doesn't decouple the $\sqrt{2-t^2}$ factor from the rest as cleanly as $t=\sin x-\cos x$ does here.
   • Sign slip when computing $(\sin x+\cos x)^2$ vs $(\sin x-\cos x)^2$ and which one equals $1\pm2\sin x\cos x$.

   ✓Final answer

   The correct option is (D) — $(2\sqrt{2},\sin x-\cos x)$.

   ANSWER: D

9. 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

10. AP EAPCET 2025Set eng-2025-05-22-FN1 markMCQ
    Q.$\int\frac{x^4 - 1}{x^2\sqrt{x^4 + x^2 + 1}}dx =$ (A) $\frac{2\sqrt{x^4+x^2+1}}{x} + c$ (B) $\frac{\sqrt{x^4+x^2+1}}{x} + c$ (C) $\frac{\sqrt{x^4+x^2+1}}{2x} + c$ (D) $\frac{4\sqrt{x^4+x^2+1}}{x} + c$

    ›Reveal solutionSolution

    Differentiating the candidate $\frac{\sqrt{x^4+x^2+1}}{x}$ reproduces the given integrand exactly, confirming it as the antiderivative.

    Concept and Intuition

    When an integrand looks like it could come from a quotient rule (a square root over a power of $x$), it is often faster to differentiate a plausible candidate of that shape and check, rather than search for a substitution from scratch.

    Step-by-Step Solution

    1. Try $g(x)=\dfrac{\sqrt{x^4+x^2+1}}{x}=\dfrac{N}{x}$ where $N=\sqrt{x^4+x^2+1}$.
    2. $N'=\dfrac{4x^3+2x}{2\sqrt{x^4+x^2+1}}=\dfrac{x(2x^2+1)}{N}$.
    3. Quotient rule: $g'(x)=\dfrac{N'x-N}{x^2}=\dfrac{\frac{x^2(2x^2+1)}{N}-N}{x^2}=\dfrac{x^2(2x^2+1)-N^2}{Nx^2}$.
    4. $N^2=x^4+x^2+1$, so the numerator is $x^2(2x^2+1)-(x^4+x^2+1)=2x^4+x^2-x^4-x^2-1=x^4-1$.
    5. So $g'(x)=\dfrac{x^4-1}{x^2\sqrt{x^4+x^2+1}}$ — exactly the given integrand.
    6. Hence $\displaystyle\int\frac{x^4-1}{x^2\sqrt{x^4+x^2+1}}\,dx=\frac{\sqrt{x^4+x^2+1}}{x}+c$.

    Common Mistakes

    • Attempting a substitution like $t=x-1/x$ or $t=x+1/x$ and getting tangled in cross terms instead of recognising the quotient-rule shape.
    • Dropping the $x^2$ in the denominator when differentiating $N/x$ (quotient rule, not just $N'/x$).

    ✓Final answer

    The correct option is (B) — $\frac{\sqrt{x^4+x^2+1}}{x} + c$.

    ANSWER: B

11. AP EAPCET 2026Set eng-2026-05-14-FN1 markMCQ
    Q.$\displaystyle\int \dfrac{dx}{(x^5+1)^{6/5}} =$ (A) $\dfrac{1}{\sqrt[5]{x^5+1}} + c$ (B) $\dfrac{\sqrt[5]{x^5+1}}{x} + c$ (C) $\dfrac{x}{\sqrt[5]{x^5+1}} + c$ (D) $\sqrt[5]{x^5+1} + c$

    ›Reveal solutionSolution

    Recognizing the integrand as the derivative of $\dfrac{x}{\sqrt[5]{x^5+1}}$ (verified by direct differentiation) gives the antiderivative immediately.

    Concept and Intuition

    For integrals of the form $\int\frac{dx}{(x^n+1)^{(n+1)/n}}$, a useful trick is to guess that the antiderivative looks like $\frac{x}{(x^n+1)^{1/n}}$ (a ratio designed to make the product-rule differentiation collapse nicely), and then verify by differentiating it — if it reproduces the integrand exactly, we're done. This is often faster than a substitution for this particular family.

    Step-by-Step Solution

    1. Guess the antiderivative $g(x) = \dfrac{x}{(x^5+1)^{1/5}} = x(x^5+1)^{-1/5}$.
    2. Differentiate using the product rule: $g'(x) = (x^5+1)^{-1/5} + x\cdot\left(-\dfrac15\right)(x^5+1)^{-6/5}\cdot5x^4$.
    3. Simplify the second term: $x\cdot\left(-\dfrac15\right)(5x^4)(x^5+1)^{-6/5} = -x^5(x^5+1)^{-6/5}$.
    4. So $g'(x) = (x^5+1)^{-1/5} - x^5(x^5+1)^{-6/5}$.
    5. Factor out $(x^5+1)^{-6/5}$: $g'(x) = (x^5+1)^{-6/5}\left[(x^5+1) - x^5\right] = (x^5+1)^{-6/5}\times 1 = (x^5+1)^{-6/5}$.
    6. This matches the integrand $\dfrac{1}{(x^5+1)^{6/5}}$ exactly.
    7. So $\displaystyle\int\dfrac{dx}{(x^5+1)^{6/5}} = \dfrac{x}{(x^5+1)^{1/5}}+c = \dfrac{x}{\sqrt[5]{x^5+1}}+c$.

    Common Mistakes

    • Forgetting the chain rule factor of $5x^4$ when differentiating $(x^5+1)^{-1/5}$-type expressions.
    • Choosing the wrong candidate antiderivative form (e.g. without the $x$ in the numerator), which would not reproduce the integrand upon differentiation.

    ✓Final answer

    The correct option is (C) — $\dfrac{x}{\sqrt[5]{x^5+1}} + c$.

    ANSWER: C

12. AP EAPCET 2025Set eng-2025-05-21-FN1 markMCQ
    Q.If $\displaystyle\int \dfrac{dx}{\sin^3 x+\cos^3 x} = A\log\left|\dfrac{\sqrt{2}+t}{\sqrt{2}-t}\right|+B\mathrm{Tan}^{-1}(t)+c$, then $\left(\dfrac{B}{A},t\right) =$ (A) $(3\sqrt{2},\sin x-\cos x)$ (B) $(2\sqrt{2},\sin x-\cos x)$ (C) $\left(\dfrac{\sqrt{2}}{3},\sin x-\cos x\right)$ (D) $\left(\dfrac{3}{\sqrt{2}},\sin x+\cos x\right)$

    ›Reveal solutionSolution

    The standard $\sin^3x+\cos^3x$ integral, solved via the substitution $t=\sin x-\cos x$; matching to the given $A\log|\cdots|+B\tan^{-1}t$ form gives $B/A=2\sqrt2$ with $t=\sin x-\cos x$.

    Concept and Intuition

    $\sin^3x+\cos^3x$ factors as a sum of cubes, $(\sin x+\cos x)(\sin^2x-\sin x\cos x+\cos^2x)=(\sin x+\cos x)(1-\sin x\cos x)$. Both remaining factors can be expressed in terms of $t=\sin x-\cos x$ (since $t^2=1-2\sin x\cos x$ links $\sin x\cos x$ to $t$, and $dt=(\sin x+\cos x)\,dx$ conveniently cancels the other factor).

    Step-by-Step Solution

    1. Factor: $\sin^3x+\cos^3x = (\sin x+\cos x)(1-\sin x\cos x)$.
    2. Let $t=\sin x-\cos x \Rightarrow dt = (\cos x+\sin x)\,dx$, and $t^2 = 1-2\sin x\cos x \Rightarrow \sin x\cos x = \dfrac{1-t^2}{2}$.
    3. So $1-\sin x\cos x = 1-\dfrac{1-t^2}{2} = \dfrac{1+t^2}{2}$.
    4. Also $(\sin x+\cos x)^2 = 1+2\sin x\cos x = 1+(1-t^2) = 2-t^2$, so $\sin x+\cos x=\sqrt{2-t^2}$.
    5. Rewrite the integral:

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

    Since $dx = \dfrac{dt}{\sin x+\cos x} = \dfrac{dt}{\sqrt{2-t^2}}$:

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

    6. Split using $1=\dfrac{(2-t^2)+(1+t^2)}{3}$:

    $\frac{2}{(2-t^2)(1+t^2)} = \frac23\cdot\frac{(2-t^2)+(1+t^2)}{(2-t^2)(1+t^2)} = \frac23\left[\frac{1}{1+t^2}+\frac{1}{2-t^2}\right].$

    7. Integrate each piece: $\displaystyle\int\frac{dt}{1+t^2}=\tan^{-1}t$; $\displaystyle\int\frac{dt}{2-t^2}=\frac{1}{2\sqrt2}\log\left|\frac{\sqrt2+t}{\sqrt2-t}\right|$ (standard form $\int\frac{dx}{a^2-x^2}=\frac{1}{2a}\log\left|\frac{a+x}{a-x}\right|$ with $a=\sqrt2$).
    8. So the integral is

    $\frac23\tan^{-1}t + \frac23\cdot\frac{1}{2\sqrt2}\log\left|\frac{\sqrt2+t}{\sqrt2-t}\right|+c = \frac{1}{3\sqrt2}\log\left|\frac{\sqrt2+t}{\sqrt2-t}\right| + \frac23\tan^{-1}t + c.$

    9. Matching to $A\log\left|\frac{\sqrt2+t}{\sqrt2-t}\right|+B\tan^{-1}t+c$: $A=\dfrac{1}{3\sqrt2}$, $B=\dfrac23$.
    10. $\dfrac{B}{A} = \dfrac{2/3}{1/(3\sqrt2)} = \dfrac23\times3\sqrt2 = 2\sqrt2$, and $t=\sin x-\cos x$.

    Common Mistakes

    • Using $t=\sin x+\cos x$ instead of $\sin x-\cos x$ (the sign matters — it's the difference that makes $dt$ match the other factor cleanly here).
    • Errors in the standard log-form constant $\dfrac{1}{2a}$ with $a=\sqrt2$, or forgetting the extra $\frac23$ scaling from the partial-fraction split.

    ✓Final answer

    The correct option is (B) — $(2\sqrt2,\ \sin x-\cos x)$.

    ANSWER: B