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

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

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integral of 1/((x+4)(x-1))

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

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

Solution steps

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

Take the partial fraction of

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

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

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

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

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

Add a constant to the solution

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

y = e^{x}, y = e^{- 3 x}, x = ln (5)

\int x^{2} + 2 x + 5 d x

f (x) = x^{2} + 8, (4, 24)

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

\int x^{4} sin (3 x) d x

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