How do you find #int (2x)/((1-x)(1+x^2)) dx# using partial fractions?

Question

Ezekiel Alexander · Accepted Answer

#int (2xdx)/((1-x)(1+x^2)) =-ln abs(1-x)+1/2ln(1+x^2)-arctanx#

Develop the integral in partial fractions:

#(2x)/((1-x)(1+x^2)) = A/(1-x) + (Bx+C)/(1+x^2)#

#(2x)/((1-x)(1+x^2)) = (A(1+x^2) + (Bx+C)(1-x))/((1-x)(1+x^2))#

#2x = A+Ax^2 +Bx+C-Bx^2-Cx#

#2x = (A-B)x^2 +(B-C)x+(A+C)#

So:

# A-B = 0 => A=B#

#B-C = 2 => A-C=2#

#[(A-C=2) , (A+C = 0)] => A=1, B=1, C=-1#

#(2x)/((1-x)(1+x^2)) = 1/(1-x) +(x-1)/(1+x^2)#

#int (2xdx)/((1-x)(1+x^2)) = int (dx)/(1-x) +int (xdx)/(1+x^2)-int (dx)/(1+x^2)=-ln abs(1-x)+1/2ln(1+x^2)-arctanx#

Emilia Aguilar · Answer

undefined To find ∫(2x)/((1-x)(1+x^2)) dx using partial fractions, follow these steps:

1. Decompose the fraction into partial fractions.
2. Find the values of the unknown constants.
3. Integrate each partial fraction separately.

The partial fraction decomposition of (2x)/((1-x)(1+x^2)) is:

(2x)/((1-x)(1+x^2)) = A/(1-x) + (Bx + C)/(1+x^2)

To find the values of A, B, and C, multiply both sides by the denominator:

2x = A(1+x^2) + (Bx + C)(1-x)

Then, equate coefficients of like terms:

For x^2 term: 0 = A + B
For x term: 2 = -A + C
For constant term: 0 = A - C

Solve this system of equations to find A, B, and C.

After finding the values of A, B, and C, integrate each partial fraction separately:

∫A/(1-x) dx + ∫(Bx + C)/(1+x^2) dx

The integral of A/(1-x) with respect to x is A * ln|1-x| + C1, where C1 is the constant of integration.

The integral of (Bx + C)/(1+x^2) with respect to x can be found using trigonometric substitution or other methods depending on your preference.

This process allows you to find the integral of (2x)/((1-x)(1+x^2)) dx using partial fractions.