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 sin (3 x) sin (2 x) d x

x^{2} - 8 x + 16

\int_{2}^{\infty} \frac{1}{x ^{3}} d x

y^{^{'}} - 2 y = 3 e^{t}

15000 + 250 0^{t}

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