Q.Evaluate the definite integral $\int_{0}^{1}\left(x\,e^x+\sin\frac{\pi x}{4}\right)dx$

Showing the 12 most recent of 39 on this concept.

1. AP EAPCET 2026Set eng-2026-05-12-FN1 markMCQ
   Q.If $\displaystyle\int e^x \left(\dfrac{2-x^2}{(2+x)^{3/2}\sqrt{2-x}}\right) dx = e^x f(x) + c$, then the domain of $f(x)$ is (A) $(-\infty, -2) \cup (2, \infty)$ (B) $[-2, 2]$ (C) $(-2, 2]$ (D) $(-\infty, -2] \cup [2, \infty)$

   ›Reveal solutionSolution

   This tests the standard trick $\int e^x[f(x)+f'(x)]dx=e^xf(x)+c$; here $f(x)=\sqrt{(2-x)/(2+x)}$ and its domain is $(-2,2]$.

   Concept and Intuition

   Whenever an integral has the shape $e^x\times(\text{something})$ and the answer is stated as $e^xf(x)+c$, the 'something' must secretly be $f(x)+f'(x)$ — this is because $\dfrac{d}{dx}[e^xf(x)]=e^x[f(x)+f'(x)]$. So the real task is pattern-matching the given rational expression to a function plus its derivative.

   Step-by-Step Solution

   1. Guess a form built from the surds present: $f(x)=\sqrt{\dfrac{2-x}{2+x}}=(2-x)^{1/2}(2+x)^{-1/2}$.
   2. Differentiate using the product/chain rule: let $u=(2-x)/(2+x)$; $u'=\dfrac{-(2+x)-(2-x)}{(2+x)^2}=\dfrac{-4}{(2+x)^2}$. Then $f'(x)=\tfrac12u^{-1/2}u'=\tfrac12\sqrt{\tfrac{2+x}{2-x}}\cdot\dfrac{-4}{(2+x)^2}=\dfrac{-2}{(2+x)^{3/2}(2-x)^{1/2}}$.
   3. Add: over the common denominator $(2+x)^{3/2}(2-x)^{1/2}$, the numerator of $f(x)$ becomes $(2-x)(2+x)=4-x^2$, and $f'(x)$ contributes $-2$. Sum of numerators: $4-x^2-2=2-x^2$.
   4. So $f(x)+f'(x)=\dfrac{2-x^2}{(2+x)^{3/2}(2-x)^{1/2}}$, matching the given integrand exactly, confirming $f(x)=\sqrt{\dfrac{2-x}{2+x}}$. …

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2. AP EAPCET 2026Set eng-2026-05-13-FN1 markMCQ
   Q.If $f(x)$ is a twice differentiable function and $f'(0) = 0$, then $\displaystyle\int_0^{\pi/2} (f(x) + f''(x))\cos x\,dx =$ (A) $f\left(\dfrac{\pi}{2}\right)$ (B) $f'\left(\dfrac{\pi}{2}\right)$ (C) $1$ (D) $0$

   ›Reveal solutionSolution

   Two rounds of integration by parts on $\int_0^{\pi/2}f''(x)\cos x\,dx$ make the

   $\int f(x)\cos x\,dx$ term reappear and cancel against the same term from the

   original integral, leaving just $f(\pi/2)$.

   Concept and Intuition

   When an integral mixes a function with its own second derivative against a

   trigonometric weight, two rounds of integration by parts typically bring back a copy of

   the original integral (since differentiating $\cos x$ twice returns $-\cos x$,

   picking up sign changes along the way) — so the two copies combine or cancel, leaving

   only boundary terms.

   Step-by-Step Solution

   1. Write the target as $\displaystyle\int_0^{\pi/2}f(x)\cos x\,dx+\int_0^{\pi/2}f''(x)\cos x\,dx$.
   2. For the second integral, integrate by parts with $u=\cos x,\ dv=f''(x)\,dx$ so $du=-\sin x\,dx,\ v=f'(x)$: $\displaystyle\int_0^{\pi/2}f''(x)\cos x\,dx=\big[f'(x)\cos x\big]_0^{\pi/2} +\int_0^{\pi/2}f'(x)\sin x\,dx$.
   3. Evaluate the boundary term: $f'(\pi/2)\cos(\pi/2)-f'(0)\cos(0)=0-f'(0)\cdot1=0$ (using $\cos(\pi/2)=0$ and the given $f'(0)=0$).
   4. So $\displaystyle\int_0^{\pi/2}f''(x)\cos x\,dx=\int_0^{\pi/2}f'(x)\sin x\,dx$.
   5. Integrate this by parts again, with $u=\sin x,\ dv=f'(x)\,dx$ so $du=\cos x\,dx,\ v=f(x)$: …

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3. AP EAPCET 2026Set eng-2026-05-14-AN1 markMCQ
   Q.If $\displaystyle\int e^{x}\,\dfrac{x^3+3x^2+4}{(x+1)^3}\,dx = e^{x} f(x)+c$, then $f(x)=$ (A) $\dfrac{x^2+2x-2}{(x+1)^2}$ (B) $\dfrac{x^2+x-1}{(x+1)^2}$ (C) $\dfrac{x^2-2x+2}{(x+1)^2}$ (D) $\dfrac{x^2+2x-1}{(x+1)^2}$

   ›Reveal solutionSolution

   This is the classic $\int e^x[f(x)+f'(x)]dx=e^xf(x)+c$ recognition problem; matching the rational integrand to $g(x)+g'(x)$ for a guessed quadratic-over-$(x+1)^2$ form pins down $f(x)=\dfrac{x^2+2x-2}{(x+1)^2}$.

   Concept and Intuition

   Whenever an integral has the shape $\int e^x\cdot h(x)\,dx$ and the answer is claimed to be $e^x f(x)+c$, it must be that $h(x)=f(x)+f'(x)$ (product rule run backwards). So instead of doing a hard partial-fraction integration of a rational function times $e^x$, we can guess the shape of $f(x)$ from the options (here, degree-2 over $(x+1)^2$) and solve for its unknown coefficients algebraically.

   Step-by-Step Solution

   1. We need $f(x)$ with $f(x)+f'(x)=\dfrac{x^3+3x^2+4}{(x+1)^3}$. Try $f(x)=\dfrac{x^2+ax+b}{(x+1)^2}$.
   2. Differentiate: $f'(x)=\dfrac{(2x+a)(x+1)^2-(x^2+ax+b)\cdot2(x+1)}{(x+1)^4}=\dfrac{(2x+a)(x+1)-2(x^2+ax+b)}{(x+1)^3}$.
   3. Expand the numerator: $(2x+a)(x+1)-2(x^2+ax+b)=2x^2+(2+a)x+a-2x^2-2ax-2b=(2-a)x+(a-2b)$.
   4. So $f'(x)=\dfrac{(2-a)x+(a-2b)}{(x+1)^3}$, while $f(x)=\dfrac{(x^2+ax+b)(x+1)}{(x+1)^3}$.
   5. Add: $f(x)+f'(x)=\dfrac{(x^2+ax+b)(x+1)+(2-a)x+(a-2b)}{(x+1)^3}$.
   6. Expand $(x^2+ax+b)(x+1)=x^3+(1+a)x^2+(a+b)x+b$; adding the linear correction gives numerator $x^3+(1+a)x^2+(b+2)x+(a-b)$. …

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4. AP EAPCET 2026Set eng-2026-05-18-FN1 markMCQ
   Q.$\displaystyle\int x^2(\log x)^2\, dx =$ (A) $\dfrac{1}{6}x^3\left[3(\log x)^2 - 3\log x + 4\right] + c$ (B) $\dfrac{1}{27}x^3\left[9(\log x)^2 - 6\log x + 2\right] + c$ (C) $\dfrac{1}{27}x^3\left[9(\log x)^2 + 6\log x + 2\right] + c$ (D) $\dfrac{1}{9}x^3\left[6(\log x)^2 - 3\log x + 1\right] + c$

   ›Reveal solutionSolution

   Apply integration by parts twice (reduction formula style) on $x^2(\log x)^2$; the result is $\dfrac1{27}x^3[9(\log x)^2-6\log x+2]+c$.

   Concept and Intuition

   Each power of $\log x$ present costs one integration by parts with $u=(\log x)^k$, $dv=x^n\,dx$, reducing the power of the log by one each time. Two applications are needed here since $(\log x)^2$ appears.

   Step-by-Step Solution

   1. First application (parts, $u=(\log x)^2$, $dv=x^2\,dx$):

   $\int x^2(\log x)^2\,dx=\frac{x^3}{3}(\log x)^2-\frac{2}{3}\int x^2\log x\,dx.$

   2. Second application on $\int x^2\log x\,dx$ ($u=\log x$, $dv=x^2dx$):

   $\int x^2\log x\,dx=\frac{x^3}{3}\log x-\int\frac{x^3}{3}\cdot\frac1x\,dx=\frac{x^3}{3}\log x-\frac{x^3}{9}.$

   3. Substitute back: $\int x^2(\log x)^2\,dx=\frac{x^3}{3}(\log x)^2-\frac23\left[\frac{x^3}{3}\log x-\frac{x^3}{9}\right]=\frac{x^3}{3}(\log x)^2-\frac{2x^3}{9}\log x+\frac{2x^3}{27}.$ …

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5. AP EAPCET 2026Set eng-2026-05-18-AN1 markMCQ
   Q.If $\displaystyle\int e^x \left(\dfrac{1}{n} + \tan nx\right)\sec nx\, dx = \dfrac{1}{n}(g(x) + k) = F(x)$ and $F(0) = 1$, then $k =$ (A) $n$ (B) $n+1$ (C) $n-1$ (D) $1$

   ›Reveal solutionSolution

   Spot that the integrand is exactly the derivative of $e^x\sec(nx)/n$ (a product-rule construction), then use the initial condition $F(0)=1$ to pin down $k$.

   Concept and Intuition

   Many integrals of the form $e^x[p(x)+p'(x)]$ (or similar structured combinations) are designed to be exact derivatives of $e^x\cdot(\text{something})$, since $\dfrac{d}{dx}[e^xh(x)]=e^xh(x)+e^xh'(x)$. Recognizing $h(x)=\sec(nx)/n$ here (whose derivative is $\sec(nx)\tan(nx)$) collapses the whole integral instantly.

   Step-by-Step Solution

   1. Try $h(x)=\dfrac{\sec(nx)}{n}$. Then $h'(x)=\dfrac1n\cdot n\sec(nx)\tan(nx)=\sec(nx)\tan(nx)$.
   2. So $\dfrac{d}{dx}\left[e^xh(x)\right]=e^xh(x)+e^xh'(x)=e^x\left(\dfrac{\sec(nx)}{n}+\sec(nx)\tan(nx)\right)=e^x\left(\dfrac1n+\tan(nx)\right)\sec(nx)$ — exactly the given integrand.
   3. So $\displaystyle\int e^x\left(\dfrac1n+\tan nx\right)\sec nx\,dx=e^xh(x)+C=\dfrac{e^x\sec(nx)}{n}+C=\dfrac1n\left(e^x\sec(nx)+nC\right)$.
   4. Comparing to the given form $\dfrac1n(g(x)+k)$: $g(x)=e^x\sec(nx)$ and $k=nC$ (a constant). …

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6. AP EAPCET 2026Set eng-2026-05-18-AN1 markMCQ
   Q.If $F(x) = \displaystyle\int x(\log x)^2\, dx$ and $F(e) = \dfrac{e^2}{4}$, then $F(1) =$ (A) $0$ (B) $\dfrac{1}{4}$ (C) $\dfrac{1}{2}$ (D) $3\log\left(\dfrac{2}{e}\right)$

   ›Reveal solutionSolution

   Integrating $x(\log x)^2$ twice by parts gives a closed form; matching $F(e)$ pins the constant, then $F(1)=1/4$.

   Concept and Intuition

   $\int x(\log x)^2\,dx$ is a repeated integration-by-parts problem: each application of parts trades one power of $\log x$ for a simpler integral, since $\dfrac{d}{dx}(\log x)^2 = \dfrac{2\log x}{x}$ pairs nicely with $\int x\,dx = x^2/2$.

   Step-by-Step Solution

   1. Let $u=(\log x)^2,\ dv=x\,dx \Rightarrow du=\dfrac{2\log x}{x}dx,\ v=\dfrac{x^2}{2}$.

   $F(x)=\frac{x^2}{2}(\log x)^2-\int x\log x\,dx$

   2. For $\int x\log x\,dx$, let $u=\log x,\ dv=x\,dx$:

   $\int x\log x\,dx=\frac{x^2}{2}\log x-\frac{x^2}{4}$

   3. Substitute back:

   $F(x)=\frac{x^2}{2}(\log x)^2-\frac{x^2}{2}\log x+\frac{x^2}{4}+C$

   4. At $x=e$: $\log e=1$, so

   $F(e)=\frac{e^2}{2}-\frac{e^2}{2}+\frac{e^2}{4}+C=\frac{e^2}{4}+C$

   Given $F(e)=\dfrac{e^2}{4}$, so $C=0$. …

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7. AP EAPCET 2026Set eng-2026-05-18-AN1 markMCQ
   Q.If $\displaystyle\int e^{5x}x^n\, dx = F(n, x) + c$, then $5F(n, x) + nF(n-1, x) =$ (A) $F'(n, x) + k$ (B) $\dfrac{1}{5}F'(n, x) + k$ (C) $\dfrac{x\,F'(n,x)}{5} + k$ (D) $\dfrac{x^2 F'(n,x)}{F(n,x)} + k$

   ›Reveal solutionSolution

   Using the standard reduction formula for $\int e^{ax}x^n\,dx$ and the fact that $F'(n,x)$ is just the original integrand recovers $5F(n,x)+nF(n-1,x)=F'(n,x)$.

   Concept and Intuition

   $F(n,x)$ is defined as an antiderivative, so by the Fundamental Theorem of Calculus its derivative is simply the integrand: $F'(n,x)=e^{5x}x^n$. Separately, integration by parts on $\int e^{5x}x^n\,dx$ produces a reduction formula relating $F(n,x)$ to $F(n-1,x)$.

   Step-by-Step Solution

   1. Integrate by parts with $u=x^n,\ dv=e^{5x}dx \Rightarrow du=nx^{n-1}dx,\ v=\dfrac{e^{5x}}{5}$:

   $\int e^{5x}x^n\,dx=\frac{x^ne^{5x}}{5}-\frac{n}{5}\int e^{5x}x^{n-1}\,dx$

   2. In terms of $F$: $F(n,x)=\dfrac{x^ne^{5x}}{5}-\dfrac{n}{5}F(n-1,x)$, so

   $5F(n,x)+nF(n-1,x)=x^ne^{5x}$

   3. But since $F(n,x)+c=\int e^{5x}x^n\,dx$, differentiating both sides w.r.t. $x$ gives $F'(n,x)=e^{5x}x^n$ …

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8. AP EAPCET 2025Set eng-2025-05-21-FN1 markMCQ
   Q.$\displaystyle\int (\log x)^3 x^4\,dx =$ (A) $x^5\left[\dfrac{1}{5}(\log x)^3-\dfrac{3}{25}(\log x)^2+\dfrac{6}{125}\log x-\dfrac{6}{625}\right]+c$ (B) $x^5\left[\dfrac{1}{5}(\log x)^3-\dfrac{2}{25}(\log x)^2+\dfrac{6}{125}\log x-\dfrac{12}{125}\right]+c$ (C) $x^5\left[\dfrac{1}{5}(\log x)^3-\dfrac{4}{25}(\log x)^2-\dfrac{9}{125}\log x-\dfrac{8}{125}\right]+c$ (D) $x^5\left[\dfrac{1}{5}(\log x)^3+\dfrac{3}{25}(\log x)^2-\dfrac{6}{125}\log x-\dfrac{6}{125}\right]+c$

   ›Reveal solutionSolution

   Repeated integration by parts (equivalently, the standard reduction formula) on $\int x^4(\log x)^3\,dx$, reducing the power of $\log x$ one step at a time. Answer matches option (A).

   Concept and Intuition

   For $\int x^n(\log x)^k\,dx$, integrating by parts with $u=(\log x)^k$, $dv=x^n\,dx$ gives the reduction

   $\int x^n(\log x)^k\,dx = \frac{x^{n+1}}{n+1}(\log x)^k - \frac{k}{n+1}\int x^n(\log x)^{k-1}\,dx,$

   which is applied repeatedly until the power of $\log x$ drops to zero (a plain power integral).

   Step-by-Step Solution

   Here $n=4$, so the recursion factor is $\dfrac{k}{5}$ at each step.

   1. Base ($k=0$): $\displaystyle\int x^4\,dx = \frac{x^5}{5}$.
   2. $k=1$: $\displaystyle\int x^4\log x\,dx = \frac{x^5}{5}\log x - \frac15\int x^4\,dx = \frac{x^5}{5}\log x - \frac{x^5}{25}$.
   3. $k=2$:

   $\int x^4(\log x)^2\,dx = \frac{x^5}{5}(\log x)^2 - \frac25\int x^4\log x\,dx = \frac{x^5}{5}(\log x)^2 - \frac25\left(\frac{x^5}{5}\log x-\frac{x^5}{25}\right)$

   $= \frac{x^5}{5}(\log x)^2 - \frac{2x^5}{25}\log x + \frac{2x^5}{125}.$

   4. $k=3$:

   $\int x^4(\log x)^3\,dx = \frac{x^5}{5}(\log x)^3 - \frac35\int x^4(\log x)^2\,dx$

   $= \frac{x^5}{5}(\log x)^3 - \frac35\left(\frac{x^5}{5}(\log x)^2 - \frac{2x^5}{25}\log x + \frac{2x^5}{125}\right)$ …

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9. AP EAPCET 2025Set eng-2025-05-21-AN1 markMCQ
   Q.If $\int x^2 \cos^2 x\,dx = \dfrac{1}{6}f(x) + g(x)\sin 2x + h(x)\cos 2x + c$, then $f(1) + g(2) + h\left(\dfrac{1}{2}\right) =$ (A) $0$ (B) $2$ (C) $1$ (D) $-1$

   ›Reveal solutionSolution

   Use $\cos^2x=(1+\cos2x)/2$ to split the integral, then integrate $x^2\cos2x$ by parts (twice) to get the $\sin2x$ and $\cos2x$ coefficient functions. Answer: $f(1)+g(2)+h(1/2)=2$.

   Concept and Intuition

   Squares of trig functions are best handled via the power-reduction (double-angle) identities, turning $x^2\cos^2x$ into a polynomial term plus a polynomial-times-$\cos2x$ term. The latter is a standard repeated integration-by-parts pattern (polynomial degree 2 needs two applications), producing terms in $\sin2x$ and $\cos2x$ with polynomial coefficients — exactly the structure the question sets up.

   Step-by-Step Solution

   1. $\cos^2x = \dfrac{1+\cos2x}{2}$, so $x^2\cos^2x = \dfrac{x^2}{2} + \dfrac{x^2\cos2x}{2}$.
   2. $\displaystyle\int \frac{x^2}{2}\,dx = \frac{x^3}{6}$.
   3. For $\displaystyle\int x^2\cos2x\,dx$, integrate by parts with $u=x^2,\,dv=\cos2x\,dx$:

   $\int x^2\cos2x\,dx = \frac{x^2\sin2x}{2} - \int x\sin2x\,dx.$

   4. For $\displaystyle\int x\sin2x\,dx$, parts again with $u=x,\,dv=\sin2x\,dx$:

   $\int x\sin2x\,dx = -\frac{x\cos2x}{2} + \int \frac{\cos2x}{2}dx = -\frac{x\cos2x}{2}+\frac{\sin2x}{4}.$

   5. Combine: $\displaystyle\int x^2\cos2x\,dx = \frac{x^2\sin2x}{2} + \frac{x\cos2x}{2} - \frac{\sin2x}{4}$.
   6. Halve (from step 1's factor $\tfrac12$): $\displaystyle\int \frac{x^2\cos2x}{2}dx = \frac{x^2\sin2x}{4} + \frac{x\cos2x}{4} - \frac{\sin2x}{8}$. …

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10. AP EAPCET 2025Set eng-2025-05-23-FN1 markMCQ
    Q.$\displaystyle\int \cos^{-1}\left(\dfrac{1-x^2}{1+x^2}\right)dx=$ (A) $2[x\tan^{-1}x-\log\sqrt{1+x^2}]+c$ (B) $2x\tan^{-1}x+\log\sqrt{1+x^2}+c$ (C) $x\tan^{-1}x+\log\sqrt{1-x^2}+c$ (D) $2[\tan^{-1}x-\log\sqrt{1+x^2}]+c$

    ›Reveal solutionSolution

    Recognising the standard identity $\cos^{-1}\left(\frac{1-x^2}{1+x^2}\right)=2\tan^{-1}x$ turns this into a routine integration-by-parts problem.

    Concept and Intuition

    Substituting $x=\tan\phi$ turns $\dfrac{1-x^2}{1+x^2}$ into $\cos2\phi$ (the standard tangent half-angle relation), so $\cos^{-1}\left(\dfrac{1-x^2}{1+x^2}\right)=2\phi=2\tan^{-1}x$ (for the principal range where this holds).

    Step-by-Step Solution

    1. $\cos^{-1}\left(\dfrac{1-x^2}{1+x^2}\right)=2\tan^{-1}x$.
    2. Integral becomes $\displaystyle\int 2\tan^{-1}x\,dx$.
    3. Integrate by parts with $u=\tan^{-1}x$, $dv=dx$: $\int\tan^{-1}x\,dx=x\tan^{-1}x-\int\frac{x}{1+x^2}dx$.
    4. $\displaystyle\int\frac{x}{1+x^2}dx=\frac12\ln(1+x^2)=\log\sqrt{1+x^2}$. …

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11. AP EAPCET 2025Set eng-2025-05-26-AN1 markMCQ
    Q.$\displaystyle\int (\log 2x)^3\, dx =$ (A) $x\left[(\log 2x)^3 - 3(\log 2x)^2 + 6(\log 2x) - 6\right] + c$ (B) $\dfrac{x}{4}\left[4(\log 2x)^3 - 6(\log 2x)^2 + 6(\log 2x) - 3\right] + c$ (C) $\dfrac{x}{2}\left[(\log 2x)^3 - 3(\log 2x)^2 + 3(\log 2x) - 6\right] + c$ (D) $x\left[(\log 2x)^3 - 6(\log 2x)^2 + 18(\log 2x) - 54\right] + c$

    ›Reveal solutionSolution

    Because $\log(2x)$ differentiates exactly like $\log x$, three rounds of integration by parts reproduce the classical $(\log x)^3$ antiderivative pattern.

    Concept and Intuition

    $\dfrac{d}{dx}\log(2x)=\dfrac1x$ — the constant factor 2 inside the log vanishes on differentiation. So treating $u=\log(2x)$ behaves exactly as $u=\log x$ would under repeated integration by parts with $dv=dx$.

    Step-by-Step Solution

    1. Let $t=\log2x$. Using $\int t^3\,dx = xt^3 - 3\int t^2\,dx$ (parts: $u=t^3, dv=dx$, $du=3t^2\cdot\tfrac1x\,dx$, $v=x$).
    2. Similarly $\int t^2\,dx = xt^2 - 2\int t\,dx$, and $\int t\,dx = xt - x$.
    3. Back-substitute: $\int t^2\,dx = xt^2-2(xt-x)=xt^2-2xt+2x$.
    4. $\int t^3\,dx = xt^3-3(xt^2-2xt+2x)=xt^3-3xt^2+6xt-6x$. …

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12. AP EAPCET 2025Set eng-2025-05-26-FN1 markMCQ
    Q.$\displaystyle\int_0^1 x\,\sin^{-1}x\,dx =$ (A) $\dfrac{\pi}{8}$ (B) $\dfrac{\pi}{4}$ (C) $\dfrac{\pi}{12}$ (D) $\dfrac{\pi}{3}$

    ›Reveal solutionSolution

    Integration by parts on $\int_0^1 x\sin^{-1}x\,dx$, with the resulting $\int x^2/\sqrt{1-x^2}\,dx$ handled via a standard reduction, gives $\frac{\pi}{8}$.

    Concept and Intuition

    When the integrand is a product of a polynomial and an inverse trig function, integration by parts with $u=\sin^{-1}x$ (so $u'$ is algebraic) and $dv=x\,dx$ (easy to integrate) is the standard approach — it trades the inverse-trig factor for a simpler algebraic integral.

    Step-by-Step Solution

    1. Let $u=\sin^{-1}x$, $dv=x\,dx$. Then $du = \dfrac{dx}{\sqrt{1-x^2}}$, $v=\dfrac{x^2}{2}$.
    2. By parts: $\displaystyle\int x\sin^{-1}x\,dx = \frac{x^2}{2}\sin^{-1}x - \int \frac{x^2}{2\sqrt{1-x^2}}dx$.
    3. For $\displaystyle\int\frac{x^2}{\sqrt{1-x^2}}dx$, write $x^2 = 1-(1-x^2)$, so it equals $\displaystyle\int\frac{dx}{\sqrt{1-x^2}} - \int\sqrt{1-x^2}\,dx = \sin^{-1}x - \left[\frac{x}{2}\sqrt{1-x^2}+\frac12\sin^{-1}x\right] = \frac12\sin^{-1}x - \frac{x}{2}\sqrt{1-x^2}$.
    4. So $\displaystyle\int x\sin^{-1}x\,dx = \frac{x^2}{2}\sin^{-1}x - \frac12\left[\frac12\sin^{-1}x-\frac{x}{2}\sqrt{1-x^2}\right] = \frac{x^2}{2}\sin^{-1}x - \frac14\sin^{-1}x + \frac{x}{4}\sqrt{1-x^2}$. …

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