# Thread: Solve inverse tan integral

1. ## Solve inverse tan integral

Hey, I cannot figure out how to solve this integral. My teacher says we can't use the table of integrals in the back of the book, and have to solve it ourselves. Here it is:

$\displaystyle \int{\arctan{x}}dx$

And the answer is supposed to be:

$\displaystyle \int{\arctan{x}}dx=x*\arctan{x}-\ln{\sqrt{1+x^2}}+C$

Thanks

2. Originally Posted by DHS1
Hey, I cannot figure out how to solve this integral. My teacher says we can't use the table of integrals in the back of the book, and have to solve it ourselves. Here it is:

$\displaystyle \int{\arctan{x}}dx$

And the answer is supposed to be:

$\displaystyle \int{\arctan{x}}dx=x*\arctan{x}-\ln{\sqrt{1+x^2}}+C$

Thanks
integration by parts ...

$\displaystyle u = \arctan{x}$

$\displaystyle dv = dx$

$\displaystyle du = \frac{1}{1+x^2} \, dx$

$\displaystyle v = x$

$\displaystyle \int u \, dv = u \cdot v - \int v \, du$

take it from here?

3. Thank you skeeter that looks like it would work fine. However there is one problem, he did not teach us this method yet. It is only the 2nd week of this class, and all he has taught us was U-Substitution. Do you know of a way to solve it using this? Or is Integration by parts the only way?

4. Originally Posted by DHS1
Thank you skeeter that looks like it would work fine. However there is one problem, he did not teach us this method yet. It is only the 2nd week of this class, and all he has taught us was U-Substitution. Do you know of a way to solve it using this? Or is Integration by parts the only way?
Sorry, I do not know another method.