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#