Evaluate the following indefinite integral

Question

Evaluate the following indefinite integral:

 

∫(3x-2)/((x-1)(x²+1)) dx

Answer 1

Note that x² + 1 is an irreducible quadratic, so the partial fraction decomposition results in:

(3x-2)/[(x-1)(x²+1)] = A/(x-1) + (Bx+C)/(x²+1)

Multiplying every term by (x-1)(x²+1):

3x - 2 = A(x²+1) + (Bx+C)(x-1)

You can expand the right side then equate coefficients (probably the best technique here):

3x - 2 = Ax² + A + Bx² - Bx + Cx - C

A + B = 0, since there is no x² term on the left.

A - C = -2, comparing constants

-B + C = 3, comparing linear terms

Using the last two:

A - B = 1, then adding to the 1st:

2A = 1

A = 1/2

If A = 1/2, then C = 5/2 (from 2nd equation)

If A = 1/2, then B = -1/2 (B = -A from 1st equation).

So the original integral can be written as:

∫ [(1/2)/(x-1) + [((-1/2)x + 5/2)/(x²+1)]dx

The first term has antiderivative:

1/2 ln(|x-1|)

Separate the last two terms as follows:

[(-1/2)x / (x²+1)] + [(5/2)/(x²+1)]

For the 1st term in that last expression, let u = x²+1. the antiderivative is:

(-1/4)ln(x²+1)

For the 2nd term recognize the antiderivative of 1/(x²+1) as tan^-1(x):

(5/2)tan^-1(x)

So the general antiderivative of the original function is:

(1/2)ln(|x-1|) - (1/4)ln(x²+1) + (5/2)tan^-1(x) + C

The following Desmos graph confirms the antiderivative:

desmos.com/calculator/cyhpilrnhc