What is the integral of (x-9)/(4x^2+3x-1) ?

The integral of (x-9)/(4x^2+3x-1) is -7/4 ln|4x-1|+2ln|x+1|+C

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integral of (x-9)/(4x^2+3x-1)

\int \frac{x - 9}{4 x ^{2} + 3 x - 1} d x

- \frac{7}{4} ln ∣ 4 x - 1 ∣ + 2 ln ∣ x + 1 ∣ + C

Solution steps

\int \frac{x - 9}{4 x ^{2} + 3 x - 1} d x

Take the partial fraction of

\frac{x - 9}{4 x ^{2} + 3 x - 1} : - \frac{7}{4 x - 1} + \frac{2}{x + 1}

= \int - \frac{7}{4 x - 1} + \frac{2}{x + 1} 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{7}{4 x - 1} d x + \int \frac{2}{x + 1} d x

\int \frac{7}{4 x - 1} d x = \frac{7}{4} ln ∣ 4 x - 1 ∣

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

= - \frac{7}{4} ln ∣ 4 x - 1 ∣ + 2 ln ∣ x + 1 ∣

Add a constant to the solution

= - \frac{7}{4} ln ∣ 4 x - 1 ∣ + 2 ln ∣ x + 1 ∣ + C

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What is the integral of (x-9)/(4x^2+3x-1) ?
The integral of (x-9)/(4x^2+3x-1) is -7/4 ln|4x-1|+2ln|x+1|+C