What is the integral of (x^2-6x-5)/((x-1)^2(x^2+1)) ?

The integral of (x^2-6x-5)/((x-1)^2(x^2+1)) is 3ln|x-1|+5/(x-1)-3/2 ln|x^2+1|+3arctan(x)+C

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integral of (x^2-6x-5)/((x-1)^2(x^2+1))

\int \frac{x ^{2} - 6 x - 5}{( x - 1 ) ^{2} ( x ^{2} + 1 )} d x

3 ln ∣ x - 1 ∣ + \frac{5}{x - 1} - \frac{3}{2} ln x^{2} + 1 + 3 arctan (x) + C

Solution steps

\int \frac{x ^{2} - 6 x - 5}{( x - 1 ) ^{2} ( x ^{2} + 1 )} d x

Take the partial fraction of

\frac{x ^{2} - 6 x - 5}{( x - 1 ) ^{2} ( x ^{2} + 1 )} : \frac{3}{x - 1} - \frac{5}{( x - 1 ) ^{2}} + \frac{- 3 x + 3}{x ^{2} + 1}

= \int \frac{3}{x - 1} - \frac{5}{( x - 1 ) ^{2}} + \frac{- 3 x + 3}{x ^{2} + 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{3}{x - 1} d x - \int \frac{5}{( x - 1 ) ^{2}} d x + \int \frac{- 3 x + 3}{x ^{2} + 1} d x

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

\int \frac{5}{( x - 1 ) ^{2}} d x = - \frac{5}{x - 1}

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

= 3 ln ∣ x - 1 ∣ - (- \frac{5}{x - 1}) - \frac{3}{2} ln x^{2} + 1 + 3 arctan (x)

Simplify

= 3 ln ∣ x - 1 ∣ + \frac{5}{x - 1} - \frac{3}{2} ln x^{2} + 1 + 3 arctan (x)

Add a constant to the solution

= 3 ln ∣ x - 1 ∣ + \frac{5}{x - 1} - \frac{3}{2} ln x^{2} + 1 + 3 arctan (x) + C

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What is the integral of (x^2-6x-5)/((x-1)^2(x^2+1)) ?
The integral of (x^2-6x-5)/((x-1)^2(x^2+1)) is 3ln|x-1|+5/(x-1)-3/2 ln|x^2+1|+3arctan(x)+C