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

The integral of 1/((x^2+x)(x-1)^2) is ln|x|-1/4 ln|x+1|-3/4 ln|x-1|-1/(2(x-1))+C

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

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

ln ∣ x ∣ - \frac{1}{4} ln ∣ x + 1 ∣ - \frac{3}{4} ln ∣ x - 1 ∣ - \frac{1}{2 ( x - 1 )} + C

Solution steps

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

Take the partial fraction of

\frac{1}{( x ^{2} + x ) ( x - 1 ) ^{2}} : \frac{1}{x} - \frac{1}{4 ( x + 1 )} - \frac{3}{4 ( x - 1 )} + \frac{1}{2 ( x - 1 ) ^{2}}

= \int \frac{1}{x} - \frac{1}{4 ( x + 1 )} - \frac{3}{4 ( x - 1 )} + \frac{1}{2 ( 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}{x} d x - \int \frac{1}{4 ( x + 1 )} d x - \int \frac{3}{4 ( x - 1 )} d x + \int \frac{1}{2 ( x - 1 ) ^{2}} d x

\int \frac{1}{x} d x = ln ∣ x ∣

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

\int \frac{3}{4 ( x - 1 )} d x = \frac{3}{4} ln ∣ x - 1 ∣

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

= ln ∣ x ∣ - \frac{1}{4} ln ∣ x + 1 ∣ - \frac{3}{4} ln ∣ x - 1 ∣ - \frac{1}{2 ( x - 1 )}

Add a constant to the solution

= ln ∣ x ∣ - \frac{1}{4} ln ∣ x + 1 ∣ - \frac{3}{4} ln ∣ x - 1 ∣ - \frac{1}{2 ( x - 1 )} + C

\int (\frac{5 x ^{4} + x ^{3} - x ^{2} + x + 3}{x ^{2}}) d x

\frac{d}{d x} (sin (π cos (x)))

\int 7^{3 x} d x

\int (x + x)^{2} d x

x \to 2 lim (e^{x} - 1)

What is the integral of 1/((x^2+x)(x-1)^2) ?
The integral of 1/((x^2+x)(x-1)^2) is ln|x|-1/4 ln|x+1|-3/4 ln|x-1|-1/(2(x-1))+C