Integrating function with trigonometric identities, substitution and parts A) $\int \operatorname{cotg}^4 (3x)+1 \, dx$
B) $\int \frac{1- \sqrt{2x}}{1+ \sqrt{2x}} \, dx$
C) $\int \frac{\sqrt[3]{x^2}} {x+1} \, dx$
In A), I know that I can apply the integration sum rule, so one of the integrals is $x$. Then, in the other integral I can use u-substitution, so $u=3x$. After that, I know that $\operatorname{cotg}^4(u)= \frac{\cos^4(u)}{\sin^4(u)}$. And then? What should I do? Probably I'll have to use trig identities, but which of them and where? 
In B), I can distribute the denominator and apply the sum integration rule, but after that? If I try to use u-substitution calling $u=1+ \sqrt(2x)$, I don't know what to do with $du$...
In C), I thought that I could integrate by parts, calling $u=\sqrt[3]x^2 $ and $dv= \frac{1}{x+1}$. But then I have to integrate $Ln|x+1| \sqrt[3]x^2$, which is really messy.
 A: Hints:
A) Bioche's rules say on should use the substitution
$$t=\tan 3x,\qquad \mathrm d t=3(1+t^2)\,\mathrm dt. $$
This leads to the integral
$$\frac13\int\frac{\mathrm dt}{t^4(1+t^2)},$$
which you should decompose into partial fractions. Note that, as the fraction is even, the decomposition will have the form 
$$\frac{1}{t^4(1+t^2)}=\frac A{u^2}+\frac B{u^4}+\frac C{1+u^2}.$$
B) Use the substitution
$$u=\sqrt{2x}\iff u^2=2x, \qquad 2u\,\mathrm du=2\,\mathrm dx.$$
C) Substitution $\; t=\sqrt[3]x\iff t^3=x$, $\quad3t^2\,\mathrm dt=\mathrm dx$.
A: Hints:
Write the integrand in (B) as $$-\frac{\sqrt {2x} -1}{\sqrt{2x} +1}=-\frac{\sqrt {2x} +1-2}{\sqrt{2x} +1}=-1+\frac{2}{\sqrt{2x} +1}=-1+\frac{2/\sqrt 2}{\sqrt{x} +1/\sqrt 2}.$$ The first part can be easily done. For the second part, make $y=\sqrt x,$ so that we have $\mathrm d x=2y\mathrm d y,$ and this integral becomes $$\frac{2}{\sqrt 2}\int{\frac{2y \mathrm d y}{y +1/\sqrt 2}}=\frac{4}{\sqrt 2}\int{\frac{y+1/\sqrt 2 -1/\sqrt 2}{y +1/\sqrt 2}\mathrm d y},$$ from where you should be able to proceed.
For (C), taking $g=x^{2/3}+1,$ we obtain $\mathrm d x=\frac 32\sqrt{g-1}\mathrm d g,$ and $x=(\sqrt{g-1})^3.$ Substituting, we obtain $$\int{\frac{x^{2/3}\mathrm d x}{x+1}}=\frac 32\int {\mathrm d g},$$ from where also you can continue.
You're right that some trig identity would do the trick for (A), but I'm presently not sufficiently patient to see a way. But try to see if something comes out of $$\cot^2x=\csc^2x-1.$$
A: For the first you can use $3x=u$, and no other substitution:
$$
\cot^4u=\cot^4u+\cot^2u-\cot^2u-1+1=\cot^2u(1+\cot^2u)-(1+\cot^2u)+1
$$
so you get
$$
\int\cot^4u\,du=-\frac{\cot^3u}{3}+\cot u+u
$$
because the derivative of $\cot u$ is $-(1+\cot^2u)$.
For the second integral, use $u=1+\sqrt{2x}$, so the integral becomes
$$
\int\frac{2-u}{u}(u-1)\,du=\int\frac{3u-u^2-2}{u}\,du
$$
which is elementary.
The final one is done with $u=\sqrt[3]{x}$, so $x=u^3$ and the integral becomes
$$
\int\frac{u^2}{u^3+1}3u^2\,du
$$
that's just tedious with partial fractions.
