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

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

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

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

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

Solution steps

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

Take the partial fraction of

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

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

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

\int \frac{1}{5 ( x + 4 )} d x = \frac{1}{5} ln ∣ x + 4 ∣

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

Add a constant to the solution

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

y = (3 x^{2} + 2) (4 x - 3)

\frac{5}{s - 3} - \frac{4}{s + 6}

\frac{d}{d x} (\frac{2 x + 1}{3 - x})

\frac{1}{s ^{2} ( s + 1 )}

\frac{d ^{4}}{d x ^{4}} (arctan (x))

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