Q.Evaluate the integral using substitution $\int_{0}^{1}\frac{x}{x^2+1}\,dx$

Showing the 12 most recent of 51 on this concept.

1. 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

2. AP EAPCET 2026Set eng-2026-05-13-AN1 markMCQ
   Q.$\displaystyle\int \dfrac{dx}{x^2(x^4+1)^{3/4}}=$ (A) $\left(1+\dfrac{1}{x^4}\right)^{3/4}+c$ (B) $\left(1+\dfrac{1}{x^6}\right)^{1/2}+c$ (C) $-\left(1+\dfrac{1}{x^4}\right)^{-1/4}+c$ (D) $-\left(1+\dfrac{1}{x^4}\right)^{1/4}+c$

   ›Reveal solutionSolution

   Pulling $x^4$ out from under the radical and substituting $t=1+x^{-4}$ reduces the integral to a simple power rule, giving $-(1+1/x^4)^{1/4}+c$.

   Concept and Intuition

   When the integrand has $(x^n+1)^p$, it often helps to factor out the highest power of $x$ from inside the bracket so that a substitution like $t = 1+x^{-n}$ produces a clean differential matching the rest of the integrand.

   Step-by-Step Solution

   1. Write $(x^4+1)^{3/4} = \left(x^4\left(1+\dfrac{1}{x^4}\right)\right)^{3/4} = x^3\left(1+\dfrac{1}{x^4}\right)^{3/4}$.
   2. So $\dfrac{1}{x^2(x^4+1)^{3/4}} = \dfrac{1}{x^2\cdot x^3\left(1+\frac{1}{x^4}\right)^{3/4}} = \dfrac{1}{x^5\left(1+\frac{1}{x^4}\right)^{3/4}} = x^{-5}\left(1+x^{-4}\right)^{-3/4}$.
   3. Let $t = 1+x^{-4}$. Then $dt = -4x^{-5}dx$, so $x^{-5}dx = -\dfrac{dt}{4}$.
   4. The integral becomes $\displaystyle\int t^{-3/4}\left(-\dfrac{dt}{4}\right) = -\dfrac14\cdot\dfrac{t^{1/4}}{1/4}+c = -t^{1/4}+c$.
   5. Substituting back: $-\left(1+\dfrac{1}{x^4}\right)^{1/4}+c$.

   Common Mistakes

   • Forgetting the negative sign that comes from $dt=-4x^{-5}dx$.
   • Not factoring $x^4$ out correctly before substituting, leading to a mismatched power.

   ✓Final answer

   The correct option is (D) — $-\left(1+\dfrac{1}{x^4}\right)^{1/4}+c$.

   ANSWER: D

3. AP EAPCET 2024Set eng-2024-05-18-FN1 markMCQ
   Q.$\displaystyle\int \dfrac{1}{x^5\sqrt[5]{x^5+1}}\,dx=$ (A) $\dfrac{4}{\sqrt[5]{x^5+1}}+c$ (B) $4x^4(x^5+1)^{4/5}+c$ (C) $-\dfrac{(x^5+1)^{4/5}}{4x^4}+c$ (D) $-\dfrac{(x^5+1)^{4/5}}{4x^5}+c$

   ›Reveal solutionSolution

   Rewriting the integrand to expose $1+x^{-5}$ as the natural substitution variable solves this cleanly; the answer is $-\dfrac{(x^5+1)^{4/5}}{4x^4}+c$.

   Concept and Intuition

   When an integral mixes a power of $x$ with a root of a polynomial in $x$, factoring out the highest power of $x$ from inside the root often converts the expression into a function of $1/x$ (or $x^{-5}$ here), whose derivative is already present elsewhere in the integrand — a clean substitution.

   Step-by-Step Solution

   1. $(x^5+1)^{-1/5}=\left(x^5\left(1+x^{-5}\right)\right)^{-1/5}=x^{-1}(1+x^{-5})^{-1/5}$.
   2. So the integrand $x^{-5}(x^5+1)^{-1/5}=x^{-6}(1+x^{-5})^{-1/5}$.
   3. Let $t=1+x^{-5}$, so $dt=-5x^{-6}dx\Rightarrow x^{-6}dx=-\dfrac{dt}{5}$.
   4. Integral $=\displaystyle\int t^{-1/5}\left(-\dfrac{dt}{5}\right)=-\dfrac15\cdot\dfrac{t^{4/5}}{4/5}+c=-\dfrac14t^{4/5}+c$.
   5. Substitute back: $-\dfrac14(1+x^{-5})^{4/5}+c=-\dfrac14\left(\dfrac{x^5+1}{x^5}\right)^{4/5}+c=-\dfrac{(x^5+1)^{4/5}}{4x^4}+c$.

   Common Mistakes

   • Trying $u=x^5+1$ directly, which does not match the $x^{-5}$ factor present and leads to a messier, non-matching form.
   • Sign or exponent slip converting $x^5\cdot x^{-4}$ powers back after substitution.

   ✓Final answer

   The correct option is (C) — $-\dfrac{(x^5+1)^{4/5}}{4x^4}+c$.

   ANSWER: C

4. 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

5. AP EAPCET 2021Set eng-2021-08-25-AN1 markMCQ
   Q.$\int \cos\sqrt{x}\, dx =$ (A) $2\sqrt{x}\sin\sqrt{x} + 2\cos\sqrt{x} + c$ (B) $2\sqrt{x}\sin\sqrt{x} + 2\sin\sqrt{x} + c$ (C) $2\sqrt{x}\sin\sqrt{x} - 2\cos\sqrt{x} + c$ (D) $\sqrt{x}\cos\sqrt{x} - 2\sin\sqrt{x} + c$

   ›Reveal solutionSolution

   Substitute $t=\sqrt x$ to turn the integral into a standard integration-by-parts problem. Answer: $2\sqrt x\sin\sqrt x+2\cos\sqrt x+c$.

   Concept and Intuition

   Whenever you see $\sqrt x$ trapped inside a trig or exponential function, substituting $t=\sqrt x$ converts it into a polynomial-times-trig integral solvable by parts.

   Step-by-Step Solution

   1. Let $t=\sqrt x \Rightarrow x=t^2,\ dx=2t\,dt$.
   2. $\int\cos\sqrt x\,dx = \int\cos t\cdot 2t\,dt = 2\int t\cos t\,dt$.
   3. Integrate by parts: $\int t\cos t\,dt = t\sin t - \int\sin t\,dt = t\sin t+\cos t$.
   4. So the integral $=2(t\sin t+\cos t)+c = 2t\sin t+2\cos t+c$.
   5. Substitute back $t=\sqrt x$: $=2\sqrt x\sin\sqrt x+2\cos\sqrt x+c$.

   Common Mistakes

   • Forgetting the factor of $2t$ from $dx=2t\,dt$ when substituting.

   ✓Final answer

   The correct option is (A) — $2\sqrt x\sin\sqrt x+2\cos\sqrt x+c$.

   ANSWER: A

6. 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

7. AP EAPCET 2022Set eng-2022-07-05-FN1 markMCQ
   Q.$\displaystyle\int \dfrac{dx}{(x-3)^{4/5}(x+1)^{6/5}} =$ (A) $\dfrac{5}{4}\sqrt[5]{\dfrac{x-3}{x+1}} + C$ (B) $\dfrac{5}{4}\left(\dfrac{x+1}{x-3}\right)^{1/5} + C$ (C) $\dfrac{1}{5}\left(\dfrac{x-3}{x+1}\right)^{1/5} + C$ (D) $\dfrac{5}{4}\left(\dfrac{x-3}{x+4}\right)^{4/5} + C$

   ›Reveal solutionSolution

   Splitting off $(x+1)^2$ turns the integrand into a pure power of $t=\frac{x-3}{x+1}$, giving $\frac54\left(\frac{x-3}{x+1}\right)^{1/5}+C$.

   Concept and Intuition

   When an integrand has the form $(x-a)^{p}(x-b)^{q}$ with $p+q$ an integer (here $\frac45+\frac65=2$), factoring out $(x-b)^{p+q}$ and substituting $t=\frac{x-a}{x-b}$ collapses the whole thing to a simple power of $t$ — a standard trick for these "unequal fractional exponent" integrals.

   Step-by-Step Solution

   1. $(x-3)^{4/5}(x+1)^{6/5} = (x+1)^2\left[\dfrac{x-3}{x+1}\right]^{4/5}$ (factoring out $(x+1)^{4/5+6/5}=(x+1)^2$).
   2. So the integrand is $(x+1)^{-2}\left[\dfrac{x-3}{x+1}\right]^{-4/5}$.
   3. Let $t=\dfrac{x-3}{x+1}$. Then $\dfrac{dt}{dx} = \dfrac{(x+1)-(x-3)}{(x+1)^2} = \dfrac{4}{(x+1)^2}$, so $(x+1)^{-2}dx = \dfrac{dt}{4}$.
   4. The integral becomes $\dfrac14\displaystyle\int t^{-4/5}\,dt = \dfrac14\cdot\dfrac{t^{1/5}}{1/5}+C = \dfrac54\,t^{1/5}+C$.
   5. Substitute back: $\dfrac54\left(\dfrac{x-3}{x+1}\right)^{1/5}+C = \dfrac54\sqrt[5]{\dfrac{x-3}{x+1}}+C$.
   6. (Verified by differentiating this result — it reproduces the original integrand exactly.)

   Common Mistakes

   • Splitting the powers the wrong way (factoring out $(x-3)^2$ instead of $(x+1)^2$), which inverts the fraction inside the fifth root.
   • Losing track of the constant $5/4$ when integrating $t^{-4/5}$.

   ✓Final answer

   The correct option is (A) — $\dfrac{5}{4}\sqrt[5]{\dfrac{x-3}{x+1}} + C$.

   ANSWER: A

8. 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

9. 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

10. AP EAPCET 2025Set eng-2025-05-27-FN1 markMCQ
    Q.$\displaystyle\int \dfrac{dx}{(1 + \sqrt{x})\sqrt{x - x^2}} =$ (A) $-2\sqrt{\dfrac{1 + \sqrt{x}}{1 - \sqrt{x}}} + c$ (B) $-\sqrt{\dfrac{1 - \sqrt{x}}{1 + \sqrt{x}}} + c$ (C) $-2\sqrt{\dfrac{1 - \sqrt{x}}{1 + \sqrt{x}}} + c$ (D) $2\sqrt{\dfrac{1 + \sqrt{x}}{1 - \sqrt{x}}} + c$

    ›Reveal solutionSolution

    Substituting $t=\sqrt x$ turns the surd-heavy integrand into $\int \frac{2\,dt}{(1+t)^{3/2}(1-t)^{1/2}}$, whose antiderivative is exactly $-2\sqrt{\frac{1-t}{1+t}}$.

    Concept and Intuition

    When an integrand mixes $\sqrt x$ and $\sqrt{x-x^2}=\sqrt{x}\sqrt{1-x}$, substituting $t=\sqrt x$ clears every square root of $x$ at once, converting the whole thing into a rational-power integral in $t$ that matches the derivative of $\sqrt{\frac{1-t}{1+t}}$ — a standard "recognise the derivative" pattern worth memorising for CET-style problems.

    Step-by-Step Solution

    1. Write $\sqrt{x-x^2}=\sqrt{x(1-x)}=\sqrt x\sqrt{1-x}$, so the integral is

    $I=\int\frac{dx}{(1+\sqrt x)\sqrt x\sqrt{1-x}}.$

    2. Let $t=\sqrt x$, so $x=t^2$, $dx=2t\,dt$:

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

    3. Let $y=\sqrt{\dfrac{1-t}{1+t}}$. Differentiating $y^2=\dfrac{1-t}{1+t}$:

    $2y\,y'=\frac{-(1+t)-(1-t)}{(1+t)^2}=\frac{-2}{(1+t)^2}\ \Rightarrow\ y'=\frac{-1}{y(1+t)^2}=\frac{-1}{(1+t)^{3/2}(1-t)^{1/2}}.$

    4. So $I=2\int y'\,dt\cdot(-1)^{-1}$, i.e. $\dfrac{d}{dt}\left(-2y\right)=\dfrac{2}{(1+t)^{3/2}(1-t)^{1/2}}$, matching the integrand exactly.
    5. Hence $I=-2y+c=-2\sqrt{\dfrac{1-t}{1+t}}+c=-2\sqrt{\dfrac{1-\sqrt x}{1+\sqrt x}}+c$.

    Common Mistakes

    • Flipping the ratio inside the square root (getting $\sqrt{\frac{1+t}{1-t}}$ instead of $\sqrt{\frac{1-t}{1+t}}$) — check by differentiating your guess before committing.
    • Losing the negative sign in front.

    ✓Final answer

    The correct option is (C) — $-2\sqrt{\dfrac{1-\sqrt x}{1+\sqrt x}}+c$.

    ANSWER: C

11. 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

12. AP EAPCET 2023Set eng-2023-05-17-AN1 markMCQ
    Q.If $\displaystyle\int \dfrac{x^3}{\sqrt{1+x^2}}\,dx = A(1+x^2)^{3/2}+B(1+x^2)^{1/2}+C$, then $A+B=$ (A) $2/3$ (B) $-2/3$ (C) $1/3$ (D) $-1/3$

    ›Reveal solutionSolution

    The substitution $u=1+x^2$ turns the integral into a simple power-rule computation, giving $A=1/3$ and $B=-1$, so $A+B=-2/3$.

    Concept and Intuition

    Whenever the integrand has an odd power of $x$ alongside a function of $x^2$ (here $\sqrt{1+x^2}$), substituting $u=1+x^2$ (so $du=2x\,dx$) converts the odd-power part into a polynomial in $u$, making the integral elementary.

    Step-by-Step Solution

    1. Let $u=1+x^2$, $du=2x\,dx$, and $x^2=u-1$.
    2. $x^3\,dx=x^2\cdot x\,dx=(u-1)\cdot\dfrac{du}{2}$.
    3. $\displaystyle\int\frac{x^3}{\sqrt{1+x^2}}dx=\int\frac{(u-1)}{\sqrt u}\cdot\frac{du}{2}=\frac12\int\left(u^{1/2}-u^{-1/2}\right)du$.
    4. $=\frac12\left(\dfrac{2}{3}u^{3/2}-2u^{1/2}\right)+C=\dfrac13u^{3/2}-u^{1/2}+C$.
    5. Back-substitute: $\dfrac13(1+x^2)^{3/2}-(1+x^2)^{1/2}+C$, so $A=\dfrac13, B=-1$.
    6. $A+B=\dfrac13-1=-\dfrac23$.

    Common Mistakes

    • Losing the factor of $\tfrac12$ from $du=2x\,dx$ when substituting.
    • Sign error on $B$ (writing $B=1$ instead of $-1$).

    ✓Final answer

    The correct option is (B) — $-2/3$.

    ANSWER: B