Integration Problem

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August 2nd 2009, 02:56 PM
Ian1779

Integration Problem

Hi

I am stuck on this problem integrating the following.

$\int\frac{arctan (x)}{1+x^2}dx$

In order to integrate should I be integrating by parts and seperating the function into $arctan(x)\cdot\frac{1}{1+x^2}$

Or should I be using the substitution method to do this?

I am very stuck - I know that $\int arctan(x) = x\cdot arctan (x) - \frac {1}{2}ln(1+x^2) + c$ but don't know if it's relevant to help me solve??

Thanks
August 2nd 2009, 03:27 PM
DeMath

Quote:

Originally Posted by Ian1779

Hi

I am stuck on this problem integrating the following.

$\int\frac{arctan (x)}{1+x^2}dx$

In order to integrate should I be integrating by parts and seperating the function into $arctan(x)\cdot\frac{1}{1+x^2}$

Or should I be using the substitution method to do this?

I am very stuck - I know that $\int arctan(x) = x\cdot arctan (x) - \frac {1}{2}ln(1+x^2) + c$ but don't know if it's relevant to help me solve??

Thanks

Just substitution $\arctan x = u$

$\int {\frac{{\arctan x}}{{1 + {x^2}}}dx} = \left\{ \begin{gathered} \arctan x = u, \hfill \\\frac{{dx}}{{1 + {x^2}}} = du \hfill \\ <br /> \end{gathered} \right\} = \int {udu} = \frac{{{u^2}}}{2} + C = \frac{{{{\arctan }^2}x}}{2} + C.$
August 2nd 2009, 03:30 PM
drrevu

Use integration by parts

u=1+x^2
du=arctan(x)
v=1+x^2
dv=arctan(x)

I=integral(u*dv) -->what you're looking for

I=uv-integral(v*du)=(1+x^2)^2-integral(v*du)

But integral(v*du)=integral(u*dv) they're the same integral

so,

I=(1+x^2)^2 -I
2*I=(1+x^2)^2

I=((1+x^2)^2)/2 --->this is the answer
August 2nd 2009, 03:35 PM
Ian1779

Thanks - I never thought to substitute in for $arctan (x)$

Make sense now!!