What is the integral of (6x^2+1)/(x^2(x-1)^3) ?

The integral of (6x^2+1)/(x^2(x-1)^3) is -3ln|x|+1/x+3ln|x-1|+2/(x-1)-7/(2(x-1)^2)+C

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integral of (6x^2+1)/(x^2(x-1)^3)

\int \frac{6 x ^{2} + 1}{x ^{2} ( x - 1 ) ^{3}} d x

- 3 ln ∣ x ∣ + \frac{1}{x} + 3 ln ∣ x - 1 ∣ + \frac{2}{x - 1} - \frac{7}{2 ( x - 1 ) ^{2}} + C

Solution steps

\int \frac{6 x ^{2} + 1}{x ^{2} ( x - 1 ) ^{3}} d x

Take the partial fraction of

\frac{6 x ^{2} + 1}{x ^{2} ( x - 1 ) ^{3}} : - \frac{3}{x} - \frac{1}{x ^{2}} + \frac{3}{x - 1} - \frac{2}{( x - 1 ) ^{2}} + \frac{7}{( x - 1 ) ^{3}}

= \int - \frac{3}{x} - \frac{1}{x ^{2}} + \frac{3}{x - 1} - \frac{2}{( x - 1 ) ^{2}} + \frac{7}{( 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} d x - \int \frac{1}{x ^{2}} d x + \int \frac{3}{x - 1} d x - \int \frac{2}{( x - 1 ) ^{2}} d x + \int \frac{7}{( x - 1 ) ^{3}} d x

\int \frac{3}{x} d x = 3 ln ∣ x ∣

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

\int \frac{3}{x - 1} d x = 3 ln ∣ x - 1 ∣

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

\int \frac{7}{( x - 1 ) ^{3}} d x = - \frac{7}{2 ( x - 1 ) ^{2}}

= - 3 ln ∣ x ∣ - (- \frac{1}{x}) + 3 ln ∣ x - 1 ∣ - (- \frac{2}{x - 1}) - \frac{7}{2 ( x - 1 ) ^{2}}

Simplify

= - 3 ln ∣ x ∣ + \frac{1}{x} + 3 ln ∣ x - 1 ∣ + \frac{2}{x - 1} - \frac{7}{2 ( x - 1 ) ^{2}}

Add a constant to the solution

= - 3 ln ∣ x ∣ + \frac{1}{x} + 3 ln ∣ x - 1 ∣ + \frac{2}{x - 1} - \frac{7}{2 ( x - 1 ) ^{2}} + C

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What is the integral of (6x^2+1)/(x^2(x-1)^3) ?
The integral of (6x^2+1)/(x^2(x-1)^3) is -3ln|x|+1/x+3ln|x-1|+2/(x-1)-7/(2(x-1)^2)+C