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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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    Five answers and I am the only person who has up-voted the question.2014-07-27

5 Answers 5

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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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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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    @Matt: Long division can be used for that.2011-12-01
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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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    @tards: Right, both of them are correct.2011-12-01
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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?

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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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    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 :P2014-07-27