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

The integral of 1/((x+4)(x-1)^2) is 1/25 ln|x+4|-1/25 ln|x-1|-1/(5(x-1))+C

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

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

\frac{1}{25} ln ∣ x + 4 ∣ - \frac{1}{25} ln ∣ x - 1 ∣ - \frac{1}{5 ( x - 1 )} + C

Solution steps

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

Take the partial fraction of

\frac{1}{( x + 4 ) ( x - 1 ) ^{2}} : \frac{1}{25 ( x + 4 )} - \frac{1}{25 ( x - 1 )} + \frac{1}{5 ( x - 1 ) ^{2}}

= \int \frac{1}{25 ( x + 4 )} - \frac{1}{25 ( x - 1 )} + \frac{1}{5 ( x - 1 ) ^{2}} 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}{25 ( x + 4 )} d x - \int \frac{1}{25 ( x - 1 )} d x + \int \frac{1}{5 ( x - 1 ) ^{2}} d x

\int \frac{1}{25 ( x + 4 )} d x = \frac{1}{25} ln ∣ x + 4 ∣

\int \frac{1}{25 ( x - 1 )} d x = \frac{1}{25} ln ∣ x - 1 ∣

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

= \frac{1}{25} ln ∣ x + 4 ∣ - \frac{1}{25} ln ∣ x - 1 ∣ - \frac{1}{5 ( x - 1 )}

Add a constant to the solution

= \frac{1}{25} ln ∣ x + 4 ∣ - \frac{1}{25} ln ∣ x - 1 ∣ - \frac{1}{5 ( x - 1 )} + C

\int 5^{s i n (x)} cos (x) d x

n = 1 \sum \infty \frac{2 ^{n}}{e ^{n}}

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

\frac{d}{d x} ((2 x^{2} - 6 x + 1)^{- 8})

\int \frac{1}{2 x - 1} d x

What is the integral of 1/((x+4)(x-1)^2) ?
The integral of 1/((x+4)(x-1)^2) is 1/25 ln|x+4|-1/25 ln|x-1|-1/(5(x-1))+C