What is the integral of (-x^2-4x+18)/((3x+4)(x^2+9)) ?

The integral of (-x^2-4x+18)/((3x+4)(x^2+9)) is 2/3 ln|3x+4|-1/2 ln|x^2+9|+C

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integral of (-x^2-4x+18)/((3x+4)(x^2+9))

\int \frac{- x ^{2} - 4 x + 18}{( 3 x + 4 ) ( x ^{2} + 9 )} d x

\frac{2}{3} ln ∣ 3 x + 4 ∣ - \frac{1}{2} ln x^{2} + 9 + C

Solution steps

\int \frac{- x ^{2} - 4 x + 18}{( 3 x + 4 ) ( x ^{2} + 9 )} d x

Take the partial fraction of

\frac{- x ^{2} - 4 x + 18}{( 3 x + 4 ) ( x ^{2} + 9 )} : \frac{2}{3 x + 4} - \frac{x}{x ^{2} + 9}

= \int \frac{2}{3 x + 4} - \frac{x}{x ^{2} + 9} 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{2}{3 x + 4} d x - \int \frac{x}{x ^{2} + 9} d x

\int \frac{2}{3 x + 4} d x = \frac{2}{3} ln ∣ 3 x + 4 ∣

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

= \frac{2}{3} ln ∣ 3 x + 4 ∣ - \frac{1}{2} ln x^{2} + 9

Add a constant to the solution

= \frac{2}{3} ln ∣ 3 x + 4 ∣ - \frac{1}{2} ln x^{2} + 9 + C

What is the integral of (-x^2-4x+18)/((3x+4)(x^2+9)) ?
The integral of (-x^2-4x+18)/((3x+4)(x^2+9)) is 2/3 ln|3x+4|-1/2 ln|x^2+9|+C