Answer
$$\int\frac{18\tan^2x\sec^2x}{(2+\tan^3x)^2}dx=-\frac{6}{2+\tan^3x}+C$$
Work Step by Step
$$A=\int\frac{18\tan^2x\sec^2x}{(2+\tan^3x)^2}dx$$
a) Set $u=\tan x$, then $du=\sec^2xdx$
$$A=\int\frac{18u^2}{(2+u^3)^2}du=\int\frac{6}{(2+u^3)^2}\times3u^2du$$
Set $v=u^3$, then $dv=3u^2du$
$$A=\int\frac{6}{(2+v)^2}dv$$
Set $w=2+v$, then $dw=dv$
$$A=\int\frac{6}{w^2}dw=-6\int -\frac{1}{w^2}dw$$ $$A=-6\times\frac{1}{w}+C=-\frac{6}{w}+C$$
We have $$w=2+v=2+u^3=2+\tan^3x$$
Therefore, $$A=-\frac{6}{2+\tan^3x}+C$$
b) Set $u=\tan^3x$, then $du=3\tan^2x(\tan x)'dx=3\tan^2x\sec^2xdx$
$$A=\int\frac{6}{(2+u)^2}du$$
Set $v=2+u$, then $dv=du$
$$A=\int\frac{6}{v^2}dv=-6\int -\frac{1}{v^2}dv$$ $$A=-6\times\frac{1}{v}+C=-\frac{6}{v}+C$$
We have $$v=2+u=2+\tan^3x$$
Therefore, $$A=-\frac{6}{2+\tan^3x}+C$$
c) Set $u=2+\tan^3x$, then $du=3\tan^2x(\tan x)'dx=3\tan^2x\sec^2xdx$
$$A=\int\frac{6}{u^2}du=-6\int -\frac{1}{u^2}du$$ $$A=-6\times\frac{1}{u}+C=-\frac{6}{u}+C$$ $$A=-\frac{6}{2+\tan^3x}+C$$
