Respuesta :
[tex]\displaystyle\int\frac{1+v^2}{1-v^3}\,\mathrm dv[/tex]
[tex]1-v^3=(1-v)(1+v+v^2)[/tex]
[tex]\implies\dfrac{1+v^2}{1-v^3}=\dfrac a{1-v}+\dfrac{b_0+b_1v}{1+v+v^2}[/tex]
[tex]\implies\dfrac{1+v^2}{1-v^3}=\dfrac{a(1+v+v^2)+(b_0+b_1v)(1-v)}{1-v^3}[/tex]
[tex]\implies 1+v^2=(a+b_0)+(a-b_0+b_1)v+(a-b_1)v^2[/tex]
[tex]\implies\begin{cases}a+b_0=1\\a-b_0+b_1=0\\a-b_1=1\end{cases}\implies a=\dfrac23,b_0=\dfrac13,b_1=-\dfrac13[/tex]
So,
[tex]\displaystyle\int\frac{1+v^2}{1-v^3}\,\mathrm dv=\dfrac23\int\frac{\mathrm dv}{1-v}+\dfrac13\int\frac{1-v}{1+v+v^2}\,\mathrm dv[/tex]
The first integral is easy. For the second, since the derivative of the denominator is [tex](1+v+v^2)=1+2v[/tex], we can add and subtract [tex]3v[/tex] to get
[tex]\dfrac{1-v}{1+v+v^2}=\dfrac{1+2v-3v}{1+v+v^2}=\dfrac{1+2v}{1+v+v^2}-\dfrac{3v}{1+v+v^2}[/tex]
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:
[tex]\displaystyle\int\frac{1+2v}{1+v+v^2}\,\mathrm dv=\int\frac{\mathrm dt}t[/tex]
where [tex]t=1+v+v^2\implies\mathrm dt=(1+2v)\,\mathrm dv[/tex], and
[tex]\displaystyle\int\frac{3v}{1+v+v^2}\,\mathrm dv=3\int\frac v{\left(v+\frac12\right)^2+\frac34}\,\mathrm dv[/tex]
Then taking [tex]v+\dfrac12=\dfrac{\sqrt3}2\tan s\implies \mathrm dv=\dfrac{\sqrt3}2\sec^2s\,\mathrm ds[/tex] gives
[tex]\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[/tex]
[tex]=\displaystyle\sqrt3\int(\sqrt3\tan s-1)\,\mathrm ds[/tex]
[tex]=\displaystyle3\int\tan s\,\mathrm ds-\sqrt3\int\mathrm ds[/tex]
Now we're ready to wrap up.
[tex]\displaystyle\int\frac{1+v^2}{1-v^3}\,\mathrm dv=\dfrac23\int\frac{\mathrm dv}{1-v}+\dfrac13\int\frac{1-v}{1+v+v^2}\,\mathrm dv[/tex]
[tex]=\displaystyle-\frac23\ln|1-v|+\frac13\left(\int\frac{1+2v}{1+v+v^2}\,\mathrm dv-\int\frac{3v}{1+v+v^2}\,\mathrm dv\right)[/tex]
[tex]=\displaystyle-\frac23\ln|1-v|+\frac13\int\frac{\mathrm dt}t-\frac13\left(3\int\tan s\,\mathrm ds-\sqrt3\int\mathrm ds\right)[/tex]
[tex]=\displaystyle-\frac23\ln|1-v|+\frac13\ln|t|-\int\tan s\,\mathrm ds+\frac1{\sqrt3}\int\mathrm ds[/tex]
[tex]=\displaystyle-\frac23\ln|1-v|+\frac13\ln|1+v+v^2|+\ln|\cos s|+\frac s{\sqrt3}+C[/tex]
[tex]=\displaystyle-\frac23\ln|1-v|+\frac13\ln|1+v+v^2|+\ln\left|\frac{\sqrt3}{2\sqrt{1+v+v^2}}\right|+\frac1{\sqrt3}\tan^{-1}\frac{2v+1}{\sqrt3}+C[/tex]
This can be simplified a bit using some properties of logarithms to obtain
[tex]=\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}\frac{2v+1}{\sqrt3}+C[/tex]
[tex]=\displaystyle-\frac23\ln|1-v|-\frac16\ln(1+v+v^2)+\frac1{\sqrt3}\tan^{-1}\frac{2v+1}{\sqrt3}+C[/tex]
[tex]1-v^3=(1-v)(1+v+v^2)[/tex]
[tex]\implies\dfrac{1+v^2}{1-v^3}=\dfrac a{1-v}+\dfrac{b_0+b_1v}{1+v+v^2}[/tex]
[tex]\implies\dfrac{1+v^2}{1-v^3}=\dfrac{a(1+v+v^2)+(b_0+b_1v)(1-v)}{1-v^3}[/tex]
[tex]\implies 1+v^2=(a+b_0)+(a-b_0+b_1)v+(a-b_1)v^2[/tex]
[tex]\implies\begin{cases}a+b_0=1\\a-b_0+b_1=0\\a-b_1=1\end{cases}\implies a=\dfrac23,b_0=\dfrac13,b_1=-\dfrac13[/tex]
So,
[tex]\displaystyle\int\frac{1+v^2}{1-v^3}\,\mathrm dv=\dfrac23\int\frac{\mathrm dv}{1-v}+\dfrac13\int\frac{1-v}{1+v+v^2}\,\mathrm dv[/tex]
The first integral is easy. For the second, since the derivative of the denominator is [tex](1+v+v^2)=1+2v[/tex], we can add and subtract [tex]3v[/tex] to get
[tex]\dfrac{1-v}{1+v+v^2}=\dfrac{1+2v-3v}{1+v+v^2}=\dfrac{1+2v}{1+v+v^2}-\dfrac{3v}{1+v+v^2}[/tex]
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:
[tex]\displaystyle\int\frac{1+2v}{1+v+v^2}\,\mathrm dv=\int\frac{\mathrm dt}t[/tex]
where [tex]t=1+v+v^2\implies\mathrm dt=(1+2v)\,\mathrm dv[/tex], and
[tex]\displaystyle\int\frac{3v}{1+v+v^2}\,\mathrm dv=3\int\frac v{\left(v+\frac12\right)^2+\frac34}\,\mathrm dv[/tex]
Then taking [tex]v+\dfrac12=\dfrac{\sqrt3}2\tan s\implies \mathrm dv=\dfrac{\sqrt3}2\sec^2s\,\mathrm ds[/tex] gives
[tex]\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[/tex]
[tex]=\displaystyle\sqrt3\int(\sqrt3\tan s-1)\,\mathrm ds[/tex]
[tex]=\displaystyle3\int\tan s\,\mathrm ds-\sqrt3\int\mathrm ds[/tex]
Now we're ready to wrap up.
[tex]\displaystyle\int\frac{1+v^2}{1-v^3}\,\mathrm dv=\dfrac23\int\frac{\mathrm dv}{1-v}+\dfrac13\int\frac{1-v}{1+v+v^2}\,\mathrm dv[/tex]
[tex]=\displaystyle-\frac23\ln|1-v|+\frac13\left(\int\frac{1+2v}{1+v+v^2}\,\mathrm dv-\int\frac{3v}{1+v+v^2}\,\mathrm dv\right)[/tex]
[tex]=\displaystyle-\frac23\ln|1-v|+\frac13\int\frac{\mathrm dt}t-\frac13\left(3\int\tan s\,\mathrm ds-\sqrt3\int\mathrm ds\right)[/tex]
[tex]=\displaystyle-\frac23\ln|1-v|+\frac13\ln|t|-\int\tan s\,\mathrm ds+\frac1{\sqrt3}\int\mathrm ds[/tex]
[tex]=\displaystyle-\frac23\ln|1-v|+\frac13\ln|1+v+v^2|+\ln|\cos s|+\frac s{\sqrt3}+C[/tex]
[tex]=\displaystyle-\frac23\ln|1-v|+\frac13\ln|1+v+v^2|+\ln\left|\frac{\sqrt3}{2\sqrt{1+v+v^2}}\right|+\frac1{\sqrt3}\tan^{-1}\frac{2v+1}{\sqrt3}+C[/tex]
This can be simplified a bit using some properties of logarithms to obtain
[tex]=\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}\frac{2v+1}{\sqrt3}+C[/tex]
[tex]=\displaystyle-\frac23\ln|1-v|-\frac16\ln(1+v+v^2)+\frac1{\sqrt3}\tan^{-1}\frac{2v+1}{\sqrt3}+C[/tex]
