<s>[INST] Integration by parts with log function $$\int2x^3\ln(1-x^2)\;dx$$
How do I integrate this equation?
Thanks.
 [/INST] Use substitution rule. Let:\begin{align}u&=1-x^2\\x^2&=u+1\\du&=2x \, dx\end{align}
Then \begin{align}\int 2x^3 \ln(1-x^2) \, dx &= \int x^2\ln(1-x^2) \,2x \, dx \\
&=\int(u+1)\ln (u) \, du\end{align}
Now use integration by parts:
\begin{align}a=u+1 &\implies da=du\\
db=\ln(u) \, du &\implies b = u \ln(u)-u \end{align}
So \begin{align} \int 2x^3 \ln(1-x^2) \, dx &= \int(u+1) \ln (u) \, du \\ &= \int a \, db \\
&=ab-\int b \, da \\
&= (u+1) (u \ln(u)-u) - \int (u \ln(u)-u) \, du \\
&=(u+1)(u \ln(u)-u)- [\frac{1}{4}u^2(2 \log(u)-3)] + C \\
&=(2-x^2)[(1-x^2)\ln(1-x^2)-(1-x^2)]\\&\, \, \, \, \, \, \, \,-[\frac{1}{4}(1-x^2)^2(2\log(1-x^2)-3)]+C\end{align} 
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