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

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

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

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

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

Solution steps

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

Expand

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

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

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

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

Add a constant to the solution

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

\int sec (x) \frac{1}{x} d x

\frac{d}{d x} (\frac{x}{5} - \frac{5}{x})

\frac{\partial}{\partial p} (ln (\frac{p + q}{q ^{2}}))

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

n = 0 \sum \infty \frac{5}{e ^{2 n}}

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