# 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!

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$$\int \frac{t^3}{1+t^2} dt$$

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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ok, yea Should be able to go from there, thanks! –  Matt Dec 1 '11 at 16:26
hmm, where did that 1/2 come from? edit, nvm I see we need it to get 2dt –  Matt Dec 1 '11 at 16:31
I multiplied the denominator and the numerator by 2. See the second equation. There is a 2 in the numerator. I did this because the transformation requires that $dx = 2t dt$. –  tards Dec 1 '11 at 16:33
It will be $\frac{1}{2}(x-\ln(x))$, right? Which is $\frac{1}{2}((t^2+1)-\ln(t^2+1))$ not the one the OP posted. –  Gigili Dec 1 '11 at 17:17
You forgot a constant. The additional $\frac{1}{2}$ gets absorbed into that constant. –  tards Dec 1 '11 at 17:19
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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 $u=t^2$

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I guess im confused with that rewrite, how can you come to that? –  Matt Dec 1 '11 at 16:21
–  Quixotic Dec 1 '11 at 16:24
@Matt: Long division can be used for that. –  Ｊ. Ｍ. Dec 1 '11 at 16:28
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?