The integral from 0 to 1 of (1+2t)/(1+t^2) dt.

Thank you very much.

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- Oct 13th 2007, 03:32 PMkittycatintegration
The integral from 0 to 1 of (1+2t)/(1+t^2) dt.

Thank you very much. - Oct 13th 2007, 03:40 PMgalactus
$\displaystyle \int\frac{1+2t}{1+t^{2}}dt=\int\frac{2t}{t^{2}+1}d t+\int\frac{1}{t^{2}+1}dt$

For the first half, let $\displaystyle u=t^{2}, \;\ du=2tdt$

The other part you should recognize as arctan.