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!
 A: The first step is not correct.  You can write $\frac{t^3}{1+t^2}=t-\frac{t}{1+t^2}$  The first term integrates to $t^2/2$, the second yields to $u=t^2$
A: 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?
A: Ummmm....how did you arrive at $\frac{1}{4}t^4\left(\ln(1+t^2)\left(t+\frac{1}{3}t^3\right)\right)$?
Hint:  How about a substitution to make the denominator of $\frac{t^3}{1+t^2}$ a bit more manageable to integrate?
A: Personally, I don't like substituting too much, I prefer a more "creative" approach:
\begin{align}\int \frac{t^3}{1+t^2}dt&=\int\frac{t^3+t-t}{1+t^2}dt\\&=\int \frac{t(1+t^2)}{1+t^2}dt-\int \frac{t}{1+t^2}dt\\&=\int t\ dt-\frac{1}{2}\int \frac{2t}{1+t^2}dt\\&=\frac{t^2}{2}-\frac{1}{2}\ln(1+t^2)+C.\end{align}
I find it's easier this way, where possible.
A: $$
\int \frac{t^3}{1+t^2}\, dt = \int \frac{t^2}{1+t^2} \Big( \underbrace{{}\quad t\, dt\quad{}}_{\text{HINT}}\Big) = \int \frac{u}{1+u} \Big(\frac 1 2 \, du \Big) = \int \left( 1 - \frac{1}{1+u} \right) \Big(\frac 1 2 \, du \Big)
$$
