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

1 Answer
Oct 29, 2015

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