What is the integral of (8x+9)/((x^2+9)(8-x)) ?

The integral of (8x+9)/((x^2+9)(8-x)) is 1/2 ln|x^2+9|-ln|x-8|+C

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integral of (8x+9)/((x^2+9)(8-x))

\int \frac{8 x + 9}{( x ^{2} + 9 ) ( 8 - x )} d x

\frac{1}{2} ln x^{2} + 9 - ln ∣ x - 8 ∣ + C

Solution steps

\int \frac{8 x + 9}{( x ^{2} + 9 ) ( 8 - x )} d x

Take the partial fraction of

\frac{8 x + 9}{( x ^{2} + 9 ) ( 8 - x )} : \frac{x}{x ^{2} + 9} - \frac{1}{x - 8}

= \int \frac{x}{x ^{2} + 9} - \frac{1}{x - 8} d x

Apply the Sum Rule:

\int f (x) \pm g (x) d x = \int f (x) d x \pm \int g (x) d x

= \int \frac{x}{x ^{2} + 9} d x - \int \frac{1}{x - 8} d x

\int \frac{x}{x ^{2} + 9} d x = \frac{1}{2} ln x^{2} + 9

\int \frac{1}{x - 8} d x = ln ∣ x - 8 ∣

= \frac{1}{2} ln x^{2} + 9 - ln ∣ x - 8 ∣

Add a constant to the solution

= \frac{1}{2} ln x^{2} + 9 - ln ∣ x - 8 ∣ + C

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

What is the integral of (8x+9)/((x^2+9)(8-x)) ?
The integral of (8x+9)/((x^2+9)(8-x)) is 1/2 ln|x^2+9|-ln|x-8|+C