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Mathematics 22 Online
OpenStudy (anonymous):

Integration help! Will give a medal

OpenStudy (anonymous):

\[\int\limits_{0}^{1}\tan^{-1} \frac{ 2x-1 }{ 1-x^2+x }\]

OpenStudy (anonymous):

integration by parts maybe :) have u tried it yet?

OpenStudy (anonymous):

yeah but in vain @mukushla

OpenStudy (anonymous):

@oldrin.bataku @amistre64

OpenStudy (anonymous):

use a change of variables off the bat... \(u=1-x^2+x\) gives \(du=(-2x+1)\,dx\)

OpenStudy (anonymous):

wait are you sure you wrote it correctly?

OpenStudy (anonymous):

yeah i am correct @oldrin.bataku

OpenStudy (anonymous):

oh oops this turns into a simple result... consider \(x=t-1/2\):$$\int_0^1\arctan\frac{2x-1}{-x^2+x+1}dx=-\int_{-1/2}^{1/2}\arctan\frac{2t-2}{t^2-2t-1/4}dx$$

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