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

The integral of 1/((x-1)^2(x+4)^2) is 2/125 ln|x+4|-1/(25(x+4))-2/125 ln|x-1|-1/(25(x-1))+C

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

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

\frac{2}{125} ln ∣ x + 4 ∣ - \frac{1}{25 ( x + 4 )} - \frac{2}{125} ln ∣ x - 1 ∣ - \frac{1}{25 ( x - 1 )} + C

Solution steps

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

Apply u-substitution

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

Take the partial fraction of

\frac{1}{u ^{2} ( u - 5 ) ^{2}} : \frac{2}{125 u} + \frac{1}{25 u ^{2}} - \frac{2}{125 ( u - 5 )} + \frac{1}{25 ( u - 5 ) ^{2}}

= \int \frac{2}{125 u} + \frac{1}{25 u ^{2}} - \frac{2}{125 ( u - 5 )} + \frac{1}{25 ( u - 5 ) ^{2}} d u

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{2}{125 u} d u + \int \frac{1}{25 u ^{2}} d u - \int \frac{2}{125 ( u - 5 )} d u + \int \frac{1}{25 ( u - 5 ) ^{2}} d u

\int \frac{2}{125 u} d u = \frac{2}{125} ln ∣ u ∣

\int \frac{1}{25 u ^{2}} d u = - \frac{1}{25 u}

\int \frac{2}{125 ( u - 5 )} d u = \frac{2}{125} ln ∣ u - 5 ∣

\int \frac{1}{25 ( u - 5 ) ^{2}} d u = - \frac{1}{25 ( u - 5 )}

= \frac{2}{125} ln ∣ u ∣ - \frac{1}{25 u} - \frac{2}{125} ln ∣ u - 5 ∣ - \frac{1}{25 ( u - 5 )}

Substitute back

u = x + 4

= \frac{2}{125} ln ∣ x + 4 ∣ - \frac{1}{25 ( x + 4 )} - \frac{2}{125} ln ∣ x + 4 - 5 ∣ - \frac{1}{25 ( x + 4 - 5 )}

Simplify

\frac{2}{125} ln ∣ x + 4 ∣ - \frac{1}{25 ( x + 4 )} - \frac{2}{125} ln ∣ x + 4 - 5 ∣ - \frac{1}{25 ( x + 4 - 5 )} : \frac{2}{125} ln ∣ x + 4 ∣ - \frac{1}{25 ( x + 4 )} - \frac{2}{125} ln ∣ x - 1 ∣ - \frac{1}{25 ( x - 1 )}

= \frac{2}{125} ln ∣ x + 4 ∣ - \frac{1}{25 ( x + 4 )} - \frac{2}{125} ln ∣ x - 1 ∣ - \frac{1}{25 ( x - 1 )}

Add a constant to the solution

= \frac{2}{125} ln ∣ x + 4 ∣ - \frac{1}{25 ( x + 4 )} - \frac{2}{125} ln ∣ x - 1 ∣ - \frac{1}{25 ( x - 1 )} + C

What is the integral of 1/((x-1)^2(x+4)^2) ?
The integral of 1/((x-1)^2(x+4)^2) is 2/125 ln|x+4|-1/(25(x+4))-2/125 ln|x-1|-1/(25(x-1))+C