Substitution in an Integral from log to inverse

Question

Considering equations (2) (8) and (9) below. Please, I am wondering what is the insight into how the substitution was made to get (9). I have exhausted much energy trying to figure this out.

Note that the (2) in red is what I think was substituted into (8) from my own understanding. Also, how the $\log (1+v)$ was changed to $\frac{1}{v+1}$ is not clear to me. Pardon this seemingly easy but deep question.

Answer 1

$$C_{D2D}=\lambda_d\frac1N\int_0^\infty\log_2(1+\nu)p(\nu)d\nu\\=\lambda_d\frac1{N\ln2}\int_0^\infty \ln(1+\nu)p(\nu)d\nu$$Integrating by parts, and letting $P(\nu)=\int_\infty^\nu p(t) dt$, $$C_{D2D}=\lambda_d\frac1{N\ln2}\left[\ln(1+\nu)P(\nu)\right]_0^\infty-\lambda_d\frac1{N\ln2}\int_0^\infty \frac1{1+\nu}P(\nu)d\nu$$The boundary term vanishes, leaving $$C_{D2D}=-\lambda_d\frac1{N\ln2}\int_0^\infty \frac1{1+\nu}(-\exp(-\zeta_{dr}(\lambda_d+\gamma_{dc}\lambda_c)))d\nu$$ which is as required (though I am not too sure why the $\ln 2$ is not included in the solution in the paper).

Answer 2

Integration by parts works as follows: $$\int f'g=fg-\int fg'$$.

That would explain the appearance of $$(\log(1+\nu))'=\frac1{1+\nu}$$.