<s>[INST] Help with integrating $\int \frac{t^3}{1+t^2} ~dt$ What am I doing wrong on this integration problem?
$$
\begin{align*}
\int\frac{t^3}{1+t^2} &= 
\frac14 t^4 (\ln(1+t^2) (t+\frac13 t^3))
\\ &= \frac14 t^4(t \ln(1+t^2)+\frac13 t^3 \ln(1+t^2)
\\ &= \frac14 t^5 \ln(1+t^2)+\frac{1}{3}t^7 \ln(1+t^2)
\end{align*}$$
Answer should be $\frac{1}{2}(t^2-\ln(t^2+1))$. I'm way off
Any help appreciated.
Thanks!
 [/INST] Your integral is:
$$\int \frac{t^3}{1+t^2} dt$$
Substitute: $x = 1+t^2$ and thus $dx = 2t dt$. Then the above transforms to:
$$\int \frac{t^3}{1+t^2} dt = \frac{1}{2} \int \frac{t^2 \ 2t dt}{(1+t^2)}$$
Using the transformation suggested earlier, we can re-write the right hand side as:
$$\frac{1}{2} \int \frac{(x-1) \ dx}{x}$$
Can you take it from here?
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