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

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

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

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

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

Solution steps

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

Take the constant out:

\int a \cdot f (x) d x = a \cdot \int f (x) d x

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

Take the partial fraction of

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

= 2 \cdot \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

= 2 (- \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 ∣

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

Add a constant to the solution

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

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