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

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

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

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

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

Solution steps

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

Take the partial fraction of

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

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

\int \frac{1}{x} d x = ln ∣ x ∣

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

\int \frac{2}{x - 1} d x = 2 ln ∣ x - 1 ∣

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

Simplify

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

Add a constant to the solution

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

\int \frac{4 x + 10}{x ^{3}} d x

\int \frac{2 e ^{2 x}}{1 + e ^{4 x}} d x

\frac{d y}{d x} = 5 - 8 x

\int \frac{a ^{x}}{b ^{x}} d x

\frac{- x ^{2} - 9}{( x ^{2} - 9 ) ^{2}}

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