# Indefinite Integral Calculus II

## Homework Statement

$$\int \frac{ \tan x \sec^2 x }{ \tan^2 x + 6 \tan x + 8 } dx$$

## The Attempt at a Solution

$$\int \frac{ \tan x \sec^2 x }{ \tan^2 x + 6 \tan x + 8 } dx$$
Okay I let....

$$u=tanx$$
$$du=sec^2x$$

Then I got

$$\int \frac{ u }{ (u+4)(u+2) } du$$

Then I split this into partial fractions which is not working...... for some reason. Can somebody please help me?

$$\frac {2}{u+4} + \frac{-1}{u+2}$$