What is the integral of (x^2+2)/(x(x-1)^3) ?

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

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integral of (x^2+2)/(x(x-1)^3)

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

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

Solution steps

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

Expand

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

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

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

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

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

Simplify

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

Add a constant to the solution

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

\frac{d}{d x} (- x - \frac{16}{x})

f (x) = sin (ln (x))

\frac{d P}{d t} = P \cdot (0.0005 \cdot P - 0.1), P (0) = 300

\int x^{p} d x

\int_{2}^{4} \frac{x ^{2} ( 2 x - 3 )}{6} d x

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