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

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

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

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

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

Solution steps

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

Take the partial fraction of

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

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

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

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

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

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

Simplify

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

Add a constant to the solution

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

x \to 1 lim (\frac{3 x - 2}{2 x - 1})

\frac{d}{d x} (arctanh (3 x))

x \to 1 - lim (\frac{8}{x ^{3} - 1})

\int e^{1.2 t} d t

\frac{d}{d x} (x^{2} sin^{2} (x))

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