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

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

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

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

\frac{2}{5} ln ∣ x - 1 ∣ + \frac{3}{5} ln ∣ x + 4 ∣ + C

Solution steps

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

Take the partial fraction of

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

= \int \frac{2}{5 ( x - 1 )} + \frac{3}{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}{5 ( x - 1 )} d x + \int \frac{3}{5 ( x + 4 )} d x

\int \frac{2}{5 ( x - 1 )} d x = \frac{2}{5} ln ∣ x - 1 ∣

\int \frac{3}{5 ( x + 4 )} d x = \frac{3}{5} ln ∣ x + 4 ∣

= \frac{2}{5} ln ∣ x - 1 ∣ + \frac{3}{5} ln ∣ x + 4 ∣

Add a constant to the solution

= \frac{2}{5} ln ∣ x - 1 ∣ + \frac{3}{5} ln ∣ x + 4 ∣ + C

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