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(a) It doesn't matter to us right now that the arguments of sine and cosine are $2x$ rather than $x$. This is still the integral of powers of products of sines and cosines. Since cosine has an odd power, we make the substitution $u=\sin (2x)$, $\mathrm{d}u=2\cos(2 x)\,\mathrm{d}x$.

$$\begin{aligned} \int_0^{\pi\over 4}\sin^2(2x)\cos^3(2x)\ \mathrm{d}x &=\int_0^{\pi\over 4} \sin^2(2x)\big[1-\sin^2(2x)\big]\cos(2 x)\ \mathrm{d}x =\frac{1}{2}\int_0^1 u^2\big[1-u^2\big]\ \mathrm{d}u\\ &=\frac{1}{2}\int_0^1\big(u^2-u^4\big)\ \mathrm{d}u =\frac{1}{2}{\Big[\frac{1}{3}u^3-\frac{1}{5}u^5\Big]}_0^1 =\frac{1}{15} \end{aligned}$$

(b) Make the substitution $x=3\tan t$, $\mathrm{d}x=3\sec^2t\,\mathrm{d}t$ and use the trig identity $9+9\tan^2 t = 9\sec^2 t$.

$$\begin{aligned} \int\big(9+x^2\big)^{-{3\over 2}}\ \mathrm{d}x &=\int\big(9+9\tan^2t\big)^{-{3\over 2}}\ 3\sec^2t\,\mathrm{d}t =\int\big(3\sec t\big)^{-3}\ 3\sec^2t\,\mathrm{d}t \\ &=\frac{1}{9}\int\cos t\ \mathrm{d}t =\frac{1}{9}\sin t+C =\frac{1}{9}\frac{x}{\sqrt{x^2+9}}+C \end{aligned}$$

To convert back to $x$, in the last step, we used the triangle below, which is rigged to have $\tan t =\frac{x}{3}$.

(c) Seeing a rational function with no obvious substitutions, we use partial fractions.

$$\begin{aligned} \frac{1}{(x-1)(x^2+1)} &=\frac{a}{x-1}+\frac{bx+c}{x^2+1} =\frac{a(x^2+1)+(bx+c)(x-1)}{(x-1)(x^2+1)} \\ \text{Multiply by the original denominator.} \\ 1&=a(x^2+1)+(bx+c)(x-1)\tag{$*$} \end{aligned}$$

Setting $x=1$ gives $2a=1$ or $a=\tfrac{1}{2}$. Substituting in $a=\tfrac{1}{2}$ in $(*)$ gives

$$\begin{aligned} &\frac{1}{2}(x^2+1)+(bx+c)(x-1)=1\\ \iff\qquad& (bx+c)(x-1)=\frac{1}{2}(1-x^2)=-\frac{1}{2}(x-1)(x+1)\\ \iff\qquad& (bx+c)=-\frac{1}{2}(x+1)\\ \iff\qquad& b=c=-\frac{1}{2}\\ \end{aligned}$$

So,

$$\begin{aligned} \int\frac{\mathrm{d}x}{(x-1)(x^2+1)} &=\int\Big[\frac{1/2}{x-1}-\frac{\frac{1}{2}(x+1)}{x^2+1}\Big]\,\mathrm{d}x \\ \text{To antidifferentiate the second piece, we split it into two integrals: one that can be handled with the substitution $u=x^2+1$, and another that looks like the derivative of arctangent.} \\ &=\int\Big(\frac{1/2}{x-1}-\frac{x/2}{x^2+1} -\frac{1/2}{x^2+1}\Big)\,\mathrm{d}x\\ &=\int\Big(\frac{1/2}{x-1}-\frac{1}{4}\cdot\frac{2x}{x^2+1} -\frac{1/2}{x^2+1}\Big)\,\mathrm{d}x\\ &=\frac{1}{2}\log|x-1|-\frac{1}{4}\log(x^2+1)-\frac{1}{2}\arctan x+C \end{aligned}$$

(d) We know the derivative of arctangent, and it would integrate nicely if multiplied to the antiderivative of $x$. So, we integrate by parts with $u=\arctan x$ and $\mathrm{d}v=x\,\mathrm{d}x$ so that $\mathrm{d}u=\frac{1}{1+x^2}\mathrm{d}x$ and $v=\frac{1}{2} x^2$. Then

$$\begin{aligned} \int x\arctan x\,\mathrm{d}x&= \frac{1}{2} x^2\arctan x -\frac{1}{2}\int \frac{x^2}{1+x^2}\,\mathrm{d}x \\ &= \frac{1}{2} x^2\arctan x-\frac{1}{2}\int \frac{1+x^2-1}{1+x^2}\,\mathrm{d}x\\ &= \frac{1}{2} x^2\arctan x-\frac{1}{2}\int \left(1-\frac{1}{1+x^2}\right)\,\mathrm{d}x\\ &=\frac{1}{2}\big[ x^2\arctan x -x +\arctan x\big]+C \end{aligned}$$