Question

How do you find `int sec^2x/(tan^2x - 3tanx + 2) dx` using partial fractions?

Answer 1

`ln|(tanx-2)/(tanx-1)|+C`

Explanation:

Factor the denominator

`int(sec^2x/((tanx-1)(tanx-2)))dx`

Perform a u-substitution

Let `u = tanx` then `du=sec^2xdx`

Make the substitution into the integral

`int 1/((u-1)(u-2))du`

Now we want to do partial fraction decomposition on this

`1/((u-1)(u-2))=A/(u-1)+B/(u-2)`

`1=A(u-2)+B(u-1)`

`1=Au-2A+Bu-B`

`1=u(A+B)-2A-B`

Equating coefficients

`A+B=0` and `-2A-B=1`

Solving this system you get

`A=-1` and `B=1`

Our integral becomes

`int -1/(u-1)+1/(u-2)du`

Rewrite

`int 1/(u-2)-1/(u-1)du`

Integrating we get

`ln|tanx-2|-ln|tanx-1|+C`

Using properties of logarithms this can be rewritten as

`ln|(tanx-2)/(tanx-1)|+C`