Hi,
Can someone tell me if this correct, please?
int. tan^-1xdx=xtan^-1x-1/2ln(1+x^2)+c
I used the formula "tan^-1udu=utan^-1u-1/2ln(1+u^2)+c
Thank you
So, you can't, let's try!
By the mythic integration by parts formula, we have that
$\displaystyle \int u\,dv=u\cdot v-\int v\,du$
We gotta integrate $\displaystyle \arctan x$
So let's take $\displaystyle u=\arctan x\implies du=\frac1{1+x^2}\,dx$ & $\displaystyle dv=dx\implies v=x$, the integral becomes to
$\displaystyle \int\arctan x\,dx=x\arctan x-\int\frac x{1+x^2}\,dx$
Now the one million dollar question: could you take it from there?
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