Question

SOLVE. integration of (1+v^2) /(1-v^3)

Answer 1

`\displaystyle\int(1+v^2)/(1-v^3)\,\mathrm dv`


`1-v^3=(1-v)(1+v+v^2)`

`\implies(1+v^2)/(1-v^3)=\frac a{1-v}+(b_0+b_1v)/(1+v+v^2)`

`\implies(1+v^2)/(1-v^3)=(a(1+v+v^2)+(b_0+b_1v)(1-v))/(1-v^3)`

`\implies 1+v^2=(a+b_0)+(a-b_0+b_1)v+(a-b_1)v^2`

`\implies\begin{cases}a+b_0=1\\a-b_0+b_1=0\\a-b_1=1\end{cases}\implies a=\frac23,b_0=\frac13,b_1=-\frac13`

So,


`\displaystyle\int(1+v^2)/(1-v^3)\,\mathrm dv=\frac23\int(\mathrm dv)/(1-v)+\frac13\int(1-v)/(1+v+v^2)\,\mathrm dv`

The first integral is easy. For the second, since the derivative of the denominator is
`(1+v+v^2)=1+2v`, we can add and subtract
`3v` to get


`(1-v)/(1+v+v^2)=(1+2v-3v)/(1+v+v^2)=(1+2v)/(1+v+v^2)-(3v)/(1+v+v^2)`

and for the first term employ a substitution. For the remaining term, we can complete the square in the denominator, then use a trigonometric substitution:


`\displaystyle\int(1+2v)/(1+v+v^2)\,\mathrm dv=\int\frac{\mathrm dt}t`

where
`t=1+v+v^2\implies\mathrm dt=(1+2v)\,\mathrm dv`, and


`\displaystyle\int(3v)/(1+v+v^2)\,\mathrm dv=3\int\frac v{\left(v+\frac12\right)^2+\frac34}\,\mathrm dv`

Then taking
`v+\frac12=\frac{\sqrt3}2\tan s\implies \mathrm dv=\frac{\sqrt3}2\sec^2s\,\mathrm ds` gives


`\displaystyle3\int\frac v{\left(v+\frac12\right)^2+\frac34}\,\mathrm dv=3\int\frac{\frac{\sqrt3}2\tan s-\frac12}{\left(\frac{\sqrt3}2\tan s\right)^2+\frac34}\left(\frac{\sqrt3}2\sec^2s\right)\,\mathrm ds`

`=\displaystyle\sqrt3\int(\sqrt3\tan s-1)\,\mathrm ds`

`=\displaystyle3\int\tan s\,\mathrm ds-\sqrt3\int\mathrm ds`

Now we're ready to wrap up.


`\displaystyle\int(1+v^2)/(1-v^3)\,\mathrm dv=\frac23\int(\mathrm dv)/(1-v)+\frac13\int(1-v)/(1+v+v^2)\,\mathrm dv`

`=\displaystyle-\frac23\ln|1-v|+\frac13\left(\int(1+2v)/(1+v+v^2)\,\mathrm dv-\int(3v)/(1+v+v^2)\,\mathrm dv\right)`

`=\displaystyle-\frac23\ln|1-v|+\frac13\int\frac{\mathrm dt}t-\frac13\left(3\int\tan s\,\mathrm ds-\sqrt3\int\mathrm ds\right)`

`=\displaystyle-\frac23\ln|1-v|+\frac13\ln|t|-\int\tan s\,\mathrm ds+\frac1{\sqrt3}\int\mathrm ds`

`=\displaystyle-\frac23\ln|1-v|+\frac13\ln|1+v+v^2|+\ln|\cos s|+\frac s{\sqrt3}+C`

`=\displaystyle-\frac23\ln|1-v|+\frac13\ln|1+v+v^2|+\ln\left|(\sqrt3)/(2√(1+v+v^2))\right|+\frac1{\sqrt3}\tan^(-1)(2v+1)/(\sqrt3)+C`

This can be simplified a bit using some properties of logarithms to obtain


`=\displaystyle-\frac23\ln|1-v|+\frac13\ln(1+v+v^2)+\left(\ln\frac{\sqrt3}2-\frac12\ln(1+v+v^2)\right)+\frac1{\sqrt3}\tan^(-1)(2v+1)/(\sqrt3)+C`

`=\displaystyle-\frac23\ln|1-v|-\frac16\ln(1+v+v^2)+\frac1{\sqrt3}\tan^(-1)(2v+1)/(\sqrt3)+C`