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

Question

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

Christopher Agresta · Accepted Answer

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

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#

Axel Alvarez · Answer

undefined To find the integral of sec^2(x) / (tan^2(x) - 3tan(x) + 2) dx using partial fractions, follow these steps:

1. Factor the denominator: tan^2(x) - 3tan(x) + 2 = (tan(x) - 1)(tan(x) - 2).

2. Express the fraction as a sum of partial fractions: sec^2(x) / ((tan(x) - 1)(tan(x) - 2)) = A/(tan(x) - 1) + B/(tan(x) - 2).

3. Find the values of A and B by equating coefficients: A(tan(x) - 2) + B(tan(x) - 1) = sec^2(x).

4. Clear denominators and solve for A and B.

5. After finding A and B, integrate each term separately.

6. Finally, combine the integrals to obtain the result.