Question

Use partial fractions to find the indefinite integral. $$ \int \frac{2 x-3}{(x-1)^{2}} d x $$

Short answer

The indefinite integral \(\int \frac{2 x-3}{(x-1)^{2}} d x\) is equal to \(2 * ln|x-1| + 1/(x-1) + C \)

Step-by-step solution

Identify the decomposition types

This fraction can be decomposed as \(A/(x-1) + B/(x-1)^2\), where A and B are constants to be found.

Equate coefficients on both sides

Multiply each side by \( (x-1)^2 \) to eliminate the denominator. This gives: \(2x -3 = A*(x-1) + B\). This equation must hold for all values of x, so we can choose convenient values of x to find A and B.

Find the values of A and B

Set x = 1 to get \(2*1 - 3 = A * (1-1) + B\), which gives B = -1. Substitute x = 0 in \(2x - 3 = A*(x - 1) + B\), we get -3 = -A -1, which gives A = 2.

Rewrite the fraction

Now that we have the values of A and B, we can write the fraction as \(2/(x-1) - 1/(x-1)^2\).

Carry out the integral

Now we can calculate the integral as \(\int \frac{2}{x-1} dx - \int \frac{1}{(x-1)^2} dx\). This simplifies to \(2 * ln|x-1| + 1/(x-1) + C\), where C is the constant of integration.