Question

Integrate $\int \left(\frac{1-v^2}{v+v^3}\right)dv$.

Answer 1

$\int \left(\frac{1-v^2}{v+v^3}\right)dv$

$=-\frac{1}{3}\int \left(\frac{-3+3v^2}{v+v^3}\right)dv$

$=-\frac{1}{3}\int \frac{\left(1+3v^2\right)-4}{v+v^3}dv$

$=-\frac{1}{3}\int \frac{\left(1+3v^2\right)}{v+v^3}dv+\frac{4}{3}\int \frac{dv}{v\left(1+v^2\right)}$

putting $v+v^3=t$

$(1+3v^2)dv=dt$

$=-\frac{1}{3}\int \frac{dt}{t}+\frac{4}{3}\int \frac{dv}{v\left(1+v^2\right)}$

$=-\frac{1}{3}\log |v+v^3|+\frac{4}{3}\int \frac{dv}{v\left(1+v^2\right)}$

Let

$\frac{1}{v\left(1+v^2\right)}=\frac{A}{v}+\frac{Bv+C}{1+v^2}$

$1=A\left(1+v^2\right)+\left(Bv+C\right)v$

Put $v=0$

$A=1$

Put $v=1$

$B+C=-1$ --- (i)

put $v=-1$

$B-C=-1$ ---- (ii)

Adding (i) ad (ii)

$B=-1$ and $C=0$

So,

$\int \frac{dv}{v\left(1+v^2\right)}=\int (\frac{1}{v}+\frac{-v}{1+v^2})dv$

$\int \frac{dv}{v\left(1+v^2\right)}=\log |v|-\frac{1}{2}\log |1+v^2|$

$\int \left(\frac{1-v^2}{v+v^3}\right)dv=-\frac{1}{3}\log |v+v^3|+\frac{4}{3}\log |v|-\frac{2}{3}\log |1+v^2|+C$