What is the integral of (x+3)/((x-4)^2) ?

The integral of (x+3)/((x-4)^2) is -(x+3)/(x-4)+ln|x-4|+C

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integral of (x+3)/((x-4)^2)

\int \frac{x + 3}{( x - 4 ) ^{2}} d x

- \frac{x + 3}{x - 4} + ln ∣ x - 4 ∣ + C

Solution steps

\int \frac{x + 3}{( x - 4 ) ^{2}} d x

Apply Integration By Parts

: - \frac{x + 3}{x - 4} - \int - \frac{1}{x - 4} d x

= - \frac{x + 3}{x - 4} - \int - \frac{1}{x - 4} d x

\int - \frac{1}{x - 4} d x = - ln ∣ x - 4 ∣

= - \frac{x + 3}{x - 4} - (- ln ∣ x - 4 ∣)

Simplify

= - \frac{x + 3}{x - 4} + ln ∣ x - 4 ∣

Add a constant to the solution

= - \frac{x + 3}{x - 4} + ln ∣ x - 4 ∣ + C

\int \frac{y}{x + 1} d x

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\int \frac{1}{( 2 x + 4 ) x ^{2} + 4 x + 3} d x

\int (2 x^{5} - x^{2})^{2} (x^{\frac{- 1}{2}}) d x

What is the integral of (x+3)/((x-4)^2) ?
The integral of (x+3)/((x-4)^2) is -(x+3)/(x-4)+ln|x-4|+C