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

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

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

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

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

Solution steps

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

Expand

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

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

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

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

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

Simplify

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

Add a constant to the solution

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

\frac{d}{d x} ((3 x^{2} - 6 x + 6) e^{x})

n = 1 \sum \infty \frac{3}{4 ^{n}}

\int 3 x^{2} 1 + x^{3} d x

\int_{0}^{h} \frac{1}{2 h x} d x

\int (x^{2} + 5) ln (x) d x

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