Answer
The result is $$\int\frac{e^{3x}}{(1+e^{x})^3}dx= \ln(1+e^x)+\frac{3+4e^x}{2(1+e^x)^2}+c.$$
Work Step by Step
To integrate the integral
$$\int\frac{e^{3x}}{(1+e^{x})^3}dx$$
we will use substitution $e^x=t$ which gives us $e^xdx=dt.$
Putting this into the integral we have:
$$\int\frac{e^{3x}}{(1+e^{x})^3}dx=\int\frac{e^{2x}e^xdx}{(1+e^x)^3}=\int\frac{t^2}{(1+t)^3}dt.$$
Now we will use substitution $1+t=z\Rightarrow dt=dz$ and $t=z-1$. Putting this into integral gives us:
$$\int\frac{t^2}{(1+t)^3}dt=\int\frac{(z-1)^2}{z^3}dz=\int\frac{z^2-2z+1}{z^3}dz=\int\frac{z^2}{z^3}dz-2\int\frac{z}{z^3}dz+\int\frac{1}{z^3}dz=\int\frac{1}{z}dz-2\int\frac{1}{z^2}dz+\int\frac{1}{z^3}dz=\ln|z|-2\frac{-1}{z}-\frac{1}{2z^2}+c=\ln|z|+\frac{2}{z}-\frac{1}{2z^2}+c,$$
where $c$ is arbitrary constant. Now we have to express the solution in terms of $x$ and to do that we will express $z$ in terms of $t$ and then $t$ in terms of $x$:
$$\int\frac{e^{3x}}{(1+e^{x})^3}dx=\ln|z|+\frac{2}{z}-\frac{1}{2z^2}+c=\ln|1+t|+\frac{2}{1+t}-\frac{1}{2(1+t)^2}+c=\ln(1+e^x)+\frac{2}{1+e^x}-\frac{1}{2(1+e^x)^2}+c=\ln(1+e^x)+\frac{2\cdot2(1+e^x)}{2(1+e^x)^2}-\frac{1}{2(1+e^x)^2}+c=\ln(1+e^x)+\frac{4+4e^x-1}{2(1+e^x)^2}+c=\ln(1+e^x)+\frac{3+4e^x}{2(1+e^x)^2}+c.$$
