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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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It doesn't appear that you're doing anything right. You seem to be trying to apply a sort of chain rule that says that if $F'=f$ and $G'=g$ then $\int f(g(x))\,dx = F(g(x))G(x)+C$. That is not true. – Michael Hardy Jul 27 '14 at 12:44
Five answers and I am the only person who has up-voted the question. – Michael Hardy Jul 27 '14 at 12:44
up vote 2 down vote accepted

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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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

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
@Matt:Partial fraction decomposition. – Quixotic Dec 1 '11 at 16:24
@Matt: Long division can be used for that. – J. M. Dec 1 '11 at 16:28 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?

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$$ \int \frac{t^3}{1+t^2}\, dt = \int \frac{t^2}{1+t^2} \Big( \underbrace{{}\quad t\, dt\quad{}}_{\text{HINT}}\Big) = \int \frac{u}{1+u} \Big(\frac 1 2 \, du \Big) = \int \left( 1 - \frac{1}{1+u} \right) \Big(\frac 1 2 \, du \Big) $$

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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.

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"Personally, I don't like substituting too much", but you need substitution $u=1+t^2$ to evaluate $$ \int\frac{2t}{1+t^2}\ dt. $$ Your approach is also similar with Ross Millikan's. $\ddot\smile$ – Tunk-Fey Jul 27 '14 at 15:00
Oh, I guess you're right, that is technically a substitution. I'm so used to recognising that one that one that I don't substitute it :P – Pauly B Jul 27 '14 at 15:10

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