What is the integral of (7x+3)/(x^2+3x-4) ?

The integral of (7x+3)/(x^2+3x-4) is 2ln|x-1|+5ln|x+4|+C

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integral of (7x+3)/(x^2+3x-4)

\int \frac{7 x + 3}{x ^{2} + 3 x - 4} d x

2 ln ∣ x - 1 ∣ + 5 ln ∣ x + 4 ∣ + C

Solution steps

\int \frac{7 x + 3}{x ^{2} + 3 x - 4} d x

Take the partial fraction of

\frac{7 x + 3}{x ^{2} + 3 x - 4} : \frac{2}{x - 1} + \frac{5}{x + 4}

= \int \frac{2}{x - 1} + \frac{5}{x + 4} 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}{x - 1} d x + \int \frac{5}{x + 4} d x

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

\int \frac{5}{x + 4} d x = 5 ln ∣ x + 4 ∣

= 2 ln ∣ x - 1 ∣ + 5 ln ∣ x + 4 ∣

Add a constant to the solution

= 2 ln ∣ x - 1 ∣ + 5 ln ∣ x + 4 ∣ + C

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What is the integral of (7x+3)/(x^2+3x-4) ?
The integral of (7x+3)/(x^2+3x-4) is 2ln|x-1|+5ln|x+4|+C