What is the integral of 1/((x+1)^2(x-5)) ?

The integral of 1/((x+1)^2(x-5)) is -1/36 ln|x+1|+1/(6(x+1))+1/36 ln|x-5|+C

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integral of 1/((x+1)^2(x-5))

\int \frac{1}{( x + 1 ) ^{2} ( x - 5 )} d x

- \frac{1}{36} ln ∣ x + 1 ∣ + \frac{1}{6 ( x + 1 )} + \frac{1}{36} ln ∣ x - 5 ∣ + C

Solution steps

\int \frac{1}{( x + 1 ) ^{2} ( x - 5 )} d x

Take the partial fraction of

\frac{1}{( x + 1 ) ^{2} ( x - 5 )} : - \frac{1}{36 ( x + 1 )} - \frac{1}{6 ( x + 1 ) ^{2}} + \frac{1}{36 ( x - 5 )}

= \int - \frac{1}{36 ( x + 1 )} - \frac{1}{6 ( x + 1 ) ^{2}} + \frac{1}{36 ( x - 5 )} 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}{36 ( x + 1 )} d x - \int \frac{1}{6 ( x + 1 ) ^{2}} d x + \int \frac{1}{36 ( x - 5 )} d x

\int \frac{1}{36 ( x + 1 )} d x = \frac{1}{36} ln ∣ x + 1 ∣

\int \frac{1}{6 ( x + 1 ) ^{2}} d x = - \frac{1}{6 ( x + 1 )}

\int \frac{1}{36 ( x - 5 )} d x = \frac{1}{36} ln ∣ x - 5 ∣

= - \frac{1}{36} ln ∣ x + 1 ∣ - (- \frac{1}{6 ( x + 1 )}) + \frac{1}{36} ln ∣ x - 5 ∣

Simplify

= - \frac{1}{36} ln ∣ x + 1 ∣ + \frac{1}{6 ( x + 1 )} + \frac{1}{36} ln ∣ x - 5 ∣

Add a constant to the solution

= - \frac{1}{36} ln ∣ x + 1 ∣ + \frac{1}{6 ( x + 1 )} + \frac{1}{36} ln ∣ x - 5 ∣ + C

\frac{d}{d x} (x sin (\frac{2}{x}))

\frac{\partial}{\partial x} (\frac{- y}{2 x ^{2} + y ^{2}})

(x^{2} - 2 x) (2 x - x^{2})

\frac{\partial}{\partial t} (t e^{s t})

\frac{d}{d x} ((x - 1) tan (x) + x sin (π x))

What is the integral of 1/((x+1)^2(x-5)) ?
The integral of 1/((x+1)^2(x-5)) is -1/36 ln|x+1|+1/(6(x+1))+1/36 ln|x-5|+C