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Find the indefinite integral. $$ \int \frac{2 x}{(x-1)^{2}} d x $$

Short Answer

Expert verified
The indefinite integral of \( \int \frac{2 x}{(x-1)^{2}} dx \) is \( 2(x-1)^{-1} + 2\ln|x - 1| + C \).

Step by step solution

01

Identify substitution

Let's identify the troublesome part of the integral which is (x-1)^{2}, for simplification. Assign a new variable \( u = x - 1 \).
02

Differentiate

Differentiate this equation with respect to 'x' to get the derivative. This results in \( du = dx \).
03

Substitute in the integral

Substitute the values of 'x-1' and 'dx' back to the original integral to get a simplified version which is: \( \int \frac{2(u+1)}{u^{2}} du \).
04

Simplify integration

Split the integral and simplify further, this leads to: \( \int \frac{2u}{u^{2}} du + \int \frac{2}{u^{2}} du \). This simplifies to \( 2 \int \frac{1}{u} du - \int \frac{2}{u^{2}} du \) or \(-2\int u^{-2} du + 2 \int u^{-1} du \).
05

Integrate

Now we can correctly integrate the expressions as: \( -2 \times \left[ -u^{-1} \right] + 2 \times \left[ \ln |u| \right] \), which simplifies to \( 2u^{-1} + 2\ln|u| \).
06

Substitute 'u' back

Finally, substitute 'u' with 'x - 1' back into the equation, giving us \( 2(x-1)^{-1} + 2\ln|x - 1| + C \), where C is the constant of integration.

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