What is the integral of (x-1)/((x+3)^2) ?

The integral of (x-1)/((x+3)^2) is -(x-1)/(x+3)+ln|x+3|+C

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integral of (x-1)/((x+3)^2)

\int \frac{x - 1}{( x + 3 ) ^{2}} d x

- \frac{x - 1}{x + 3} + ln ∣ x + 3 ∣ + C

Solution steps

\int \frac{x - 1}{( x + 3 ) ^{2}} d x

Apply Integration By Parts

= - \frac{x - 1}{x + 3} - \int - \frac{1}{x + 3} d x

\int - \frac{1}{x + 3} d x = - ln ∣ x + 3 ∣

= - \frac{x - 1}{x + 3} - (- ln ∣ x + 3 ∣)

Simplify

= - \frac{x - 1}{x + 3} + ln ∣ x + 3 ∣

Add a constant to the solution

= - \frac{x - 1}{x + 3} + ln ∣ x + 3 ∣ + C

What is the integral of (x-1)/((x+3)^2) ?
The integral of (x-1)/((x+3)^2) is -(x-1)/(x+3)+ln|x+3|+C