Integral of $\int(8-\frac{7}{1+x})^2dx$

Question

What is the most elementary method for solving $\int(8-\frac{7}{1+x})^2dx$ ?

I am at a loss of how to proceed efficiently and correctly.

This happens to be the little R in a Disk-Method volume problem, with a proposed solution found here. Note: Some values are different in the solution linked.

Answer 1

\begin{align} \int(8-\frac{7}{1+x})^2dx &= \int (64 - \frac{112}{1+x} + \frac{49}{(1+x)^2}) dx \\ &= 64x -112 \ln |1+x| - \frac{49}{1+x} + C \end{align}

Answer 2

Just expand and use the linearity of the integral plus the following $$ \int\frac{1}{1+x}=\log|1+x|+C $$ and $$ \int\frac{1}{(1+x)^2}=-(1+x)^{-1}+C $$

Answer 3

Put $x+1=u$ hence $dx=du$

Hence integral turns to $$\int (8-\frac 7u)^2 du=\int 64du+\int \frac {49du}{u^2}-\int \frac {112du}{u^2}=64u-\frac {49}{u} -112\ln \vert u\vert+C=64(x+1)-\frac {49}{x+1} -112\ln \vert (x+1)\vert +C$$