Q.Evaluate the definite integral: $\int_0^1 \frac{dx}{\sqrt{1-x^2}}$

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. …

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

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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 2022Set eng-2022-07-08-FN1 markMCQ
   Q.$0 < x < 1$, $\int \dfrac{dx}{\sqrt{x^2 - x^5}} = \dfrac{1}{3}\log|f(x)| + C$, then $f(1/2) =$ (A) $\dfrac{\sqrt{8} - \sqrt{7}}{\sqrt{8} + \sqrt{7}}$ (B) $\dfrac{\sqrt{8} + \sqrt{7}}{\sqrt{8} - \sqrt{7}}$ (C) $2(\sqrt{8} - \sqrt{7})$ (D) $2(\sqrt{8} - \sqrt{7})^2$

   ›Reveal solutionSolution

   Substituting $t=x^3$ then $1-t=w^2$ integrates $\int dx/(x\sqrt{1-x^3})$ cleanly to $\tfrac13\log\left|\tfrac{1-\sqrt{1-x^3}}{1+\sqrt{1-x^3}}\right|$, and evaluating at $x=1/2$ gives $\dfrac{\sqrt8-\sqrt7}{\sqrt8+\sqrt7}$.

   Concept and Intuition

   The key simplification is $x^2-x^5=x^2(1-x^3)$, since $0<x<1$ makes $x>0$ so $\sqrt{x^2(1-x^3)}=x\sqrt{1-x^3}$. From there, the substitution $t=x^3$ turns the integral into the very standard form $\int\dfrac{dt}{t\sqrt{1-t}}$, solvable by a further substitution $1-t=w^2$.

   Step-by-Step Solution

   1. Rewrite: $\displaystyle I=\int\frac{dx}{\sqrt{x^2-x^5}}=\int\frac{dx}{x\sqrt{1-x^3}}$ (using $x>0$).
   2. Let $t=x^3\Rightarrow dt=3x^2dx\Rightarrow dx=\dfrac{dt}{3x^2}$. Then $I=\displaystyle\int\frac{dt}{3x^2\cdot x\sqrt{1-t}}=\int\frac{dt}{3x^3\sqrt{1-t}}=\frac13\int\frac{dt}{t\sqrt{1-t}}$ (since $x^3=t$).
   3. Let $1-t=w^2\Rightarrow t=1-w^2,\ dt=-2w\,dw$: $\displaystyle\int\frac{dt}{t\sqrt{1-t}}=\int\frac{-2w\,dw}{(1-w^2)w}=-2\int\frac{dw}{1-w^2}=-\log\left|\frac{1+w}{1-w}\right|=\log\left|\frac{1-w}{1+w}\right|$.
   4. So $I=\dfrac13\log\left|\dfrac{1-w}{1+w}\right|+C$ where $w=\sqrt{1-t}=\sqrt{1-x^3}$, i.e. $f(x)=\dfrac{1-\sqrt{1-x^3}}{1+\sqrt{1-x^3}}$.
   5. Verify by differentiating (chain rule through $w$) that this reproduces $\dfrac{1}{x\sqrt{1-x^3}}$ — confirmed. …

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5. AP EAPCET 2023Set eng-2023-05-18-FN1 markMCQ
   Q.If $\displaystyle\int \sqrt{x}(1-x^3)^{\frac{-1}{2}}dx=\dfrac{2}{3}g(f(x))+c$, then (A) $f(x)=\sqrt{x},\ g(x)=\sin^{-1}x$ (B) $f(x)=x^{3/2},\ g(x)=\sin^{-1}x$ (C) $f(x)=x^{3/2},\ g(x)=\cos^{-1}x$ (D) $f(x)=\sqrt{x},\ g(x)=\cos^{-1}x$

   ›Reveal solutionSolution

   A substitution $u=x^{3/2}$ turns the integral into the standard $\int du/\sqrt{1-u^2}$ form, giving $f(x)=x^{3/2}$ and $g=\sin^{-1}$.

   Concept and Intuition

   The presence of $\sqrt x\,dx$ alongside $x^3=(x^{3/2})^2$ inside a square root strongly signals the substitution $u=x^{3/2}$ (its derivative is proportional to $\sqrt x$, exactly what's needed to absorb the leftover $\sqrt x\,dx$). Once substituted, the integral collapses to the standard arcsine form.

   Step-by-Step Solution

   1. Let $u=x^{3/2}$. Then $du=\dfrac32 x^{1/2}\,dx=\dfrac32\sqrt x\,dx$, so $\sqrt x\,dx=\dfrac23\,du$.
   2. Also, $u^2=x^3$, so $1-x^3=1-u^2$.
   3. Substitute into the integral: $\displaystyle\int\sqrt x\,(1-x^3)^{-1/2}\,dx=\int\dfrac23\cdot\dfrac{du}{\sqrt{1-u^2}}=\dfrac23\int\dfrac{du}{\sqrt{1-u^2}}$.
   4. This is the standard form: $\displaystyle\int\dfrac{du}{\sqrt{1-u^2}}=\sin^{-1}u+c$.
   5. So the integral $=\dfrac23\sin^{-1}(u)+c=\dfrac23\sin^{-1}\left(x^{3/2}\right)+c$. …

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6. AP EAPCET 2025Set eng-2025-05-22-AN1 markMCQ
   Q.If $\displaystyle\int \dfrac{\cos^3 x}{\sin^2 x + \sin^4 x}dx = c - \mathrm{cosec}\,x - f(x)$, then $f\left(\dfrac{\pi}{2}\right)=$ (A) 1 (B) 0 (C) $\dfrac{\pi}{2}$ (D) $\pi$

   ›Reveal solutionSolution

   Substituting $s=\sin x$ and partial-fractioning gives $-\mathrm{cosec}\,x-2\mathrm{Tan}^{-1}(\sin x)$, so $f(x)=2\mathrm{Tan}^{-1}(\sin x)$ and $f(\pi/2)=\pi/2$.

   Concept and Intuition

   Writing $\cos^3x\,dx = \cos^2x\cdot\cos x\,dx=(1-\sin^2x)\,d(\sin x)$ converts a trig integral into an algebraic one in $s=\sin x$, which is then handled by ordinary partial fractions.

   Step-by-Step Solution

   1. Let $s=\sin x$, so $ds=\cos x\,dx$, and $\cos^3x\,dx=(1-s^2)\,ds$.
   2. The integral becomes $\displaystyle\int\frac{1-s^2}{s^2+s^4}\,ds=\int\frac{1-s^2}{s^2(1+s^2)}\,ds$.
   3. Partial fractions (in $u=s^2$): $\dfrac{1-u}{u(1+u)}=\dfrac1u-\dfrac2{1+u}$, so the integrand is $\dfrac1{s^2}-\dfrac2{1+s^2}$.
   4. Integrating: $\displaystyle\int\left(\frac1{s^2}-\frac2{1+s^2}\right)ds=-\frac1s-2\,\mathrm{Tan}^{-1}s+C=-\mathrm{cosec}\,x-2\,\mathrm{Tan}^{-1}(\sin x)+C$. …

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7. AP EAPCET 2024Set eng-2024-05-21-FN1 markMCQ
   Q.If $n \ge 2$ is a natural number and $0 < \theta < \dfrac{\pi}{2}$, then $\displaystyle\int \dfrac{(\cos^n\theta - \cos\theta)^{1/n}}{\cos^{n+1}\theta}\sin\theta\, d\theta =$ (A) $\dfrac{n}{n-1}\left(\cos^{(1-n)}\theta - 1\right)^2 + c$ (B) $\dfrac{n}{(n+1)(1-n)}\left(\cos^{(1-n)}\theta - 1\right)^{1+\frac{1}{n}} + c$ (C) $\dfrac{1}{n-1}\left(\cos^{(n-1)}\theta - 1\right)^2 + c$ (D) $\dfrac{n}{1-n^2}\left(1 - \cos^{(1-n)}\theta\right)^{(n+1)/n}$

   ›Reveal solutionSolution

   Factor out $\cos^n\theta$ from inside the radical to isolate a clean power of $\cos\theta$, then substitute $w=\cos^{1-n}\theta$ to reduce the whole integral to $\int w^{1/n}dw$.

   Concept and Intuition

   The key move is factoring $\cos^n\theta-\cos\theta=\cos^n\theta\left(1-\cos^{1-n}\theta\right)$ so the $n$-th root pulls a clean $\cos\theta$ outside, cancelling nicely against the $\cos^{n+1}\theta$ in the denominator and leaving a single power of $w=\cos^{1-n}\theta$ whose differential exactly matches $\sin\theta\,d\theta/\cos^n\theta$ in the integrand.

   Step-by-Step Solution

   1. Factor: $\cos^n\theta-\cos\theta=\cos^n\theta\left(1-\cos^{1-n}\theta\right)$, so

   $(\cos^n\theta-\cos\theta)^{1/n}=\cos\theta\left(1-\cos^{1-n}\theta\right)^{1/n}.$

   2. Divide by $\cos^{n+1}\theta$:

   $\frac{(\cos^n\theta-\cos\theta)^{1/n}}{\cos^{n+1}\theta}=\frac{\left(1-\cos^{1-n}\theta\right)^{1/n}}{\cos^n\theta}.$

   3. Let $w=\cos^{1-n}\theta$. Then $\dfrac{dw}{d\theta}=(1-n)\cos^{-n}\theta\cdot(-\sin\theta)=(n-1)\sin\theta\cos^{-n}\theta$, so $\dfrac{\sin\theta\,d\theta}{\cos^n\theta}=\dfrac{dw}{n-1}$.
   4. The whole integrand times $d\theta$ becomes $(1-w)^{1/n}\cdot\dfrac{dw}{n-1}$ — wait, more directly $w^{1/n}$ is the factor $\left(1-\cos^{1-n}\theta\right)^{1/n}$ once we track $1-\cos^{1-n}\theta$ as the base; carrying the substitution through consistently: …

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8. 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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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 $x=\pi/6$: $2x=\pi/3$, $\sin(\pi/3)=\dfrac{\sqrt3}{2}$. …

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10. 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: $\int t\cos t\,dt = t\sin t - \int\sin t\,dt = t\sin t+\cos t$. …

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11. 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 $x^{-5}dx = -\dfrac{dt}{4}$. …

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12. 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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