Q.Evaluate the definite integral: $\int_{0}^{1} \frac{2x+3}{5x^2+1} \, dx$

Showing the 12 most recent of 51 on this concept.

1. 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: $\dfrac{t^2-3}{t^2-1}=\dfrac{(t^2-1)-2}{t^2-1}=1-\dfrac{2}{t^2-1}$. …

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2. 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 $\dfrac14\displaystyle\int t^{-4/5}\,dt = \dfrac14\cdot\dfrac{t^{1/5}}{1/5}+C = \dfrac54\,t^{1/5}+C$. …

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3. 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 …

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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 $(x^5+1)^{-6/5}$: $g'(x) = (x^5+1)^{-6/5}\left[(x^5+1) - x^5\right] = (x^5+1)^{-6/5}\times 1 = (x^5+1)^{-6/5}$. …

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5. 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. …

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6. AP EAPCET 2025Set eng-2025-05-23-AN1 markMCQ
   Q.If $\displaystyle\int \dfrac{dx}{2\cos x + 3\sin x + 4} = \dfrac{2}{\sqrt{3}}f(x) + c$, then $f\left(\dfrac{2\pi}{3}\right) =$ (A) $\dfrac{\pi}{12}$ (B) $\dfrac{\pi}{8}$ (C) $\dfrac{5\pi}{12}$ (D) $\dfrac{5\pi}{8}$

   ›Reveal solutionSolution

   This is a Weierstrass ($t=\tan(x/2)$) substitution problem for a linear combination of sine and cosine plus a constant in the denominator. Evaluating $f$ at the given point gives $\dfrac{5\pi}{12}$, option (C).

   Concept and Intuition

   Whenever the denominator mixes $\sin x$, $\cos x$ and a constant, the universal substitution $t=\tan(x/2)$ (with $\cos x=\frac{1-t^2}{1+t^2}$, $\sin x=\frac{2t}{1+t^2}$, $dx=\frac{2\,dt}{1+t^2}$) converts the trigonometric denominator into a plain quadratic in $t$, reducing the whole problem to a standard $\int\frac{dt}{\text{quadratic}}$ that integrates to an arctangent.

   Step-by-Step Solution

   1. Substitute: denominator becomes

   $2\cdot\frac{1-t^2}{1+t^2}+3\cdot\frac{2t}{1+t^2}+4=\frac{2-2t^2+6t+4+4t^2}{1+t^2}=\frac{2t^2+6t+6}{1+t^2}.$

   2. The integral becomes $\displaystyle\int\frac{2\,dt/(1+t^2)}{(2t^2+6t+6)/(1+t^2)}=\int\frac{2\,dt}{2t^2+6t+6}=\int\frac{dt}{t^2+3t+3}$.
   3. Complete the square: $t^2+3t+3=\left(t+\tfrac32\right)^2+\tfrac34$, so $\int\frac{dt}{\left(t+\frac32\right)^2+\frac34}=\frac{1}{\sqrt{3}/2}\arctan\!\left(\frac{t+3/2}{\sqrt3/2}\right)+c=\frac{2}{\sqrt3}\arctan\!\left(\frac{2t+3}{\sqrt3}\right)+c.$ …

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7. 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 $=\displaystyle\int t^{-1/5}\left(-\dfrac{dt}{5}\right)=-\dfrac15\cdot\dfrac{t^{4/5}}{4/5}+c=-\dfrac14t^{4/5}+c$. …

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8. 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. $=\frac12\left(\dfrac{2}{3}u^{3/2}-2u^{1/2}\right)+C=\dfrac13u^{3/2}-u^{1/2}+C$. …

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9. AP EAPCET 2026Set eng-2026-05-15-AN1 markMCQ
   Q.If $\int x^5 e^{-4x^3}dx = \frac{1}{48}e^{-4x^3}f(x) + c$, then $f(x) =$ (A) $-2x^3 - 1$ (B) $-4x^3 - 1$ (C) $-2x^2 + 1$ (D) $4x^3 + 1$

   ›Reveal solutionSolution

   Substituting $u=x^3$ converts the integral into a simple integration-by-parts problem $\int ue^{-4u}du$; matching the result to the given form yields $f(x)=-4x^3-1$.

   Concept and Intuition

   The presence of $x^5$ alongside $e^{-4x^3}$ is a strong hint to substitute $u=x^3$, since then $x^2\,dx$ (part of $du$) combines with the remaining $x^3=u$ to leave a clean polynomial-times-exponential integral, solvable by the standard integration-by-parts reduction formula for $\int u\,e^{ku}du$.

   Step-by-Step Solution

   1. Let $u=x^3$, so $du=3x^2\,dx \Rightarrow x^2dx=\dfrac{du}{3}$.
   2. Rewrite $x^5e^{-4x^3}dx = x^3\cdot x^2 e^{-4x^3}dx = u\,e^{-4u}\cdot\dfrac{du}{3}$.
   3. So the integral becomes $\dfrac13\displaystyle\int u\,e^{-4u}\,du$.
   4. Integrate by parts with first function $u$, second $e^{-4u}$: $\int u e^{-4u}du = u\cdot\left(-\dfrac{e^{-4u}}{4}\right) - \int\left(-\dfrac{e^{-4u}}{4}\right)du = -\dfrac{u}{4}e^{-4u} - \dfrac{1}{16}e^{-4u}$. …

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10. 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 $\displaystyle\int_4^5\left(\frac{-3}{t-2}+\frac{4}{t-3}\right)dt=\Big[-3\log|t-2|+4\log|t-3|\Big]_4^5$. …

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11. AP EAPCET 2025Set eng-2025-05-21-AN1 markMCQ
    Q.If $\displaystyle\int \dfrac{dx}{(x-1)^{3/2}(x-3)^{1/2}} = \sqrt{f(x)} + c$ then $f(-1) - f(0) =$ (A) $-3$ (B) $-4$ (C) $-2$ (D) $-1$

    ›Reveal solutionSolution

    A classic "divide by $(x-1)^2$ and substitute $t=(x-3)/(x-1)$" integral. It reduces to $\sqrt{t}$, giving $f(x)=(x-3)/(x-1)$, and then $f(-1)-f(0)=-1$.

    Concept and Intuition

    When an integrand has two linear factors under different fractional powers whose exponents sum to an integer (here $3/2+1/2=2$), factor out $(x-1)^2$ from the denominator and express the rest as a function of the ratio $\dfrac{x-3}{x-1}$. This ratio then becomes a natural single substitution variable.

    Step-by-Step Solution

    1. Write the denominator as

    $(x-1)^{3/2}(x-3)^{1/2} = (x-1)^2\cdot\frac{(x-3)^{1/2}}{(x-1)^{1/2}} = (x-1)^2\sqrt{\frac{x-3}{x-1}}.$

    2. So the integral is $\displaystyle\int \frac{dx}{(x-1)^2\sqrt{\dfrac{x-3}{x-1}}}$.
    3. Let $t = \dfrac{x-3}{x-1} = 1 - \dfrac{2}{x-1}$. Then $\dfrac{dt}{dx} = \dfrac{2}{(x-1)^2}$, i.e. $\dfrac{dx}{(x-1)^2} = \dfrac{dt}{2}$.
    4. The integral becomes $\displaystyle\int \frac{dt/2}{\sqrt t} = \frac12\cdot 2\sqrt t + c = \sqrt t + c = \sqrt{\frac{x-3}{x-1}} + c$. …

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12. 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 $a=3$: $\int\dfrac{dt}{9-t^2}=\dfrac16\log\left|\dfrac{3+t}{3-t}\right|+C$. …

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