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

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

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

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

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

Solution steps

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

Expand

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

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

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

= 3 (- \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

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

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

Add a constant to the solution

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

\int x (x + 5)^{4} d x

\int cos^{(3)} (5 x) sin^{- 2} (5 x) d x

\frac{\partial}{\partial x} (e^{x^{5}})

\frac{\partial}{\partial x} (\frac{x}{( 1 - x ) ^{2}})

\frac{\partial}{\partial y} (\frac{1}{x ^{3} - y ^{2}})

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