How would you integrate

(2+t)^2 / 1+t^2 dt

Had a long uni break and now i'm trying to get back into it but i've forgotten a lot.

Please show intermediate steps. I've tried integration by parts, is that the right way to tackle it?

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- Feb 27th 2009, 02:40 PMGigabyteIntegration Help
How would you integrate

(2+t)^2 / 1+t^2 dt

Had a long uni break and now i'm trying to get back into it but i've forgotten a lot.

Please show intermediate steps. I've tried integration by parts, is that the right way to tackle it? - Feb 27th 2009, 03:02 PMo_O
Note that:

$\displaystyle \begin{aligned}(2+t)^2 & = t^2 + 4t + 4 \\ & = {\color{red}t^2 + 1} + 4t + 3 \end{aligned}$

So: $\displaystyle \frac{(2+t)^2}{1+t^2} = \frac{{\color{red}t^2 + 1} + 4t + 3 }{1 + t^2} = \underbrace{\frac{{\color{red} t^2 + 1}}{1+t^2}}_{=1} + \frac{4t }{1+t^2} + \frac{3}{1+t^2}$

So your integral becomes:

$\displaystyle \int \frac{(2+t)^2}{1+t^2} \ dt \ \ = \ \ \int \left( 1 + \frac{4t}{1+t^2} + 3 \cdot \frac{1}{1+t^2}\right) dt$

On the RHS, use a $\displaystyle u$-sub for the second integral and $\displaystyle \arctan (x)$ should ring a bell for the third. - Feb 27th 2009, 03:56 PMGigabyteThanks mate
Thanks for that, the world needs more people like yourself.

Just did it and got

2 ln|1+t^2| + 3tan^-1(t) + t + C