Integrate the following with respect to $x$:

(i) $\dfrac{x\sin^{-1}x}{\sqrt{1-x^{2}}}$

(ii) $x^{5}e^{x^{2}}$

(iii) $ an^{-1}\left(\dfrac{8x}{1-16x^{2}} ight)$

(iv) $\sin^{-1}\left(\dfrac{2x}{1+x^{2}} ight)$

Question

Integrate the following with respect to $x$:

(i) $\dfrac{x\sin^{-1}x}{\sqrt{1-x^{2}}}$

(ii) $x^{5}e^{x^{2}}$

(iii) $	an^{-1}\left(\dfrac{8x}{1-16x^{2}}ight)$

(iv) $\sin^{-1}\left(\dfrac{2x}{1+x^{2}}ight)$

Accepted Answer

> [!TLDR]
>
> Part (i) is a by-parts step where $dv$ is chosen as the piece with a ready antiderivative; part (ii) needs a substitution before Bernoulli's formula applies; parts (iii)-(iv) simplify first via inverse-trig double-angle identities, then integrate $	an^{-1}$ by parts.

**Part (i): $\dfrac{x\sin^{-1}x}{\sqrt{1-x^2}}$.** Take $u=\sin^{-1}x$ and $dv=\dfrac{x\,dx}{\sqrt{1-x^2}}$; since $\dfrac d{dx}\left[-\sqrt{1-x^2}ight]=\dfrac{x}{\sqrt{1-x^2}}$, $v=-\sqrt{1-x^2}$, and $du=\dfrac{dx}{\sqrt{1-x^2}}$.
$\displaystyle\int\dfrac{x\sin^{-1}x}{\sqrt{1-x^2}}dx=-\sqrt{1-x^2}\sin^{-1}x+\int\sqrt{1-x^2}\cdot\dfrac{dx}{\sqrt{1-x^2}}=-\sqrt{1-x^2}\sin^{-1}x+x+c$.
*Check:* $\dfrac d{dx}\left[x-\sqrt{1-x^2}\sin^{-1}xight]=1-\left[\dfrac{-x}{\sqrt{1-x^2}}\sin^{-1}x+\sqrt{1-x^2}\cdot\dfrac1{\sqrt{1-x^2}}ight]=1-1+\dfrac{x\sin^{-1}x}{\sqrt{1-x^2}}=\dfrac{x\sin^{-1}x}{\sqrt{1-x^2}}$ ✓.

**Part (ii): $x^5e^{x^2}$.** Let $t=x^2$, $dt=2x\,dx$, so $x^5dx=x^4\cdot x\,dx=t^2\cdot\dfrac{dt}2$: $\displaystyle\int x^5e^{x^2}dx=\dfrac12\int t^2e^t\,dt$.
Apply Bernoulli to $\displaystyle\int t^2e^t\,dt$ with $u=t^2\,(u'=2t,\,u''=2)$, $dv=e^tdt\,(v=e^t,\,v_1=e^t,\,v_2=e^t)$: $\displaystyle\int t^2e^t\,dt=t^2e^t-2te^t+2e^t+c$.
So $\dfrac12\int t^2e^t\,dt=\dfrac{t^2e^t}2-te^t+e^t+c$; re-substituting $t=x^2$: $\dfrac{x^4e^{x^2}}2-x^2e^{x^2}+e^{x^2}+c$.
*Check:* Differentiating term by term and collecting gives $x^5e^{x^2}$ (the $\pm2x^3e^{x^2}$ and $\pm2xe^{x^2}$ pieces cancel, leaving only $x^5e^{x^2}$) ✓.

**Part (iii): $	an^{-1}\!\left(\dfrac{8x}{1-16x^2}ight)$.** With $u=4x$, $2u=8x$ and $u^2=16x^2$, so by the identity $	an^{-1}\!\dfrac{2u}{1-u^2}=2	an^{-1}u$: the integrand is $2	an^{-1}(4x)$.
By parts on $\displaystyle\int	an^{-1}(4x)\,dx$: $u=	an^{-1}(4x)\,(du=	frac4{1+16x^2}dx)$, $dv=dx\,(v=x)$: $\displaystyle\int	an^{-1}(4x)dx=x	an^{-1}(4x)-4\int\dfrac{x\,dx}{1+16x^2}$.
For the remaining integral, put $w=1+16x^2,\,dw=32x\,dx$: $4\displaystyle\int\dfrac{x\,dx}{1+1

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