# Thread: How to integrate when the inverse tangent is present?

1. ## How to integrate when the inverse tangent is present?

Ok, one of my homework problems for my calculus class is the following:

integral (0.5 to 1) [(x^2tan^-1(x) + 3x + 1)/ x^4 + x^2] dx

here is a link to a picture of in case it doesn't make sense the way i typed it.
http://img11.imageshack.us/img11/8223/mathfp6.jpg

I really don't know how exactly to solve it. I have never done any where the inverse tangent is on the numerator. Can I get some help?

2. Is...

$\int_{.5}^{1} \frac{x^{2}\cdot \tan^{-1} x + 3 x + 1}{x^{4} + x^{2}} \cdot dx = \int_{.5}^{1} \frac{\tan^{-1} x}{1+x^{2}}\cdot dx + \int_{.5}^{1} \frac{1 + 3 x}{x^{2}\cdot (1+x^{2})}$ (1)

The integral that contains $\tan^{-1} x$ is relatively easy to solve because is...

$\frac{d}{dx} \tan^{-1} x = \frac{1}{1 + x^{2}}$ (2)

Kind regards

$\chi$ $\sigma$