Q.Integrate the following with respect to $x$:

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 $\dfrac d{dx}\left[-\sqrt{1-x^2}\right]=\dfrac{x}{\sqrt{1-x^2}}$, $v=-\sqrt{1-x^2}$, 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: $\dfrac d{dx}\left[x-\sqrt{1-x^2}\sin^{-1}x\right]=1-\left[\dfrac{-x}{\sqrt{1-x^2}}\sin^{-1}x+\sqrt{1-x^2}\cdot\dfrac1{\sqrt{1-x^2}}\right]=1-1+\dfrac{x\sin^{-1}x}{\sqrt{1-x^2}}=\dfrac{x\sin^{-1}x}{\sqrt{1-x^2}}$ ✓.

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): $\tan^{-1}\!\left(\dfrac{8x}{1-16x^2}\right)$. With $u=4x$, $2u=8x$ and $u^2=16x^2$, so by the identity $\tan^{-1}\!\dfrac{2u}{1-u^2}=2\tan^{-1}u$: the integrand is $2\tan^{-1}(4x)$.

By parts on $\displaystyle\int\tan^{-1}(4x)\,dx$: $u=\tan^{-1}(4x)\,(du=\tfrac4{1+16x^2}dx)$, $dv=dx\,(v=x)$: $\displaystyle\int\tan^{-1}(4x)dx=x\tan^{-1}(4x)-4\int\dfrac{x\,dx}{1+16x^2}$.

For the remaining integral, put $w=1+16x^2,\,dw=32x\,dx$: $4\displaystyle\int\dfrac{x\,dx}{1+16x^2}=\dfrac4{32}\log(1+16x^2)=\dfrac18\log(1+16x^2)$.

So $\displaystyle\int\tan^{-1}(4x)dx=x\tan^{-1}(4x)-\dfrac18\log(1+16x^2)+c$; doubling: $2x\tan^{-1}(4x)-\dfrac14\log(1+16x^2)+c$. …

Integration by parts — non-standard $dv$, prior substitution, and inver …

• In (i), choosing $u=x$ instead of $dv=x\,dx/\sqrt{1-x^2}$, which does not simplify …