What is the integral of (x^2-8x-17)/((x-1)^2(x^2+1)) ?

The integral of (x^2-8x-17)/((x-1)^2(x^2+1)) is 9ln|x-1|+(12)/(x-1)-9/2 ln|x^2+1|+4arctan(x)+C

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integral of (x^2-8x-17)/((x-1)^2(x^2+1))

\int \frac{x ^{2} - 8 x - 17}{( x - 1 ) ^{2} ( x ^{2} + 1 )} d x

9 ln ∣ x - 1 ∣ + \frac{12}{x - 1} - \frac{9}{2} ln x^{2} + 1 + 4 arctan (x) + C

Solution steps

\int \frac{x ^{2} - 8 x - 17}{( x - 1 ) ^{2} ( x ^{2} + 1 )} d x

Take the partial fraction of

\frac{x ^{2} - 8 x - 17}{( x - 1 ) ^{2} ( x ^{2} + 1 )} : \frac{9}{x - 1} - \frac{12}{( x - 1 ) ^{2}} + \frac{- 9 x + 4}{x ^{2} + 1}

= \int \frac{9}{x - 1} - \frac{12}{( x - 1 ) ^{2}} + \frac{- 9 x + 4}{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{9}{x - 1} d x - \int \frac{12}{( x - 1 ) ^{2}} d x + \int \frac{- 9 x + 4}{x ^{2} + 1} d x

\int \frac{9}{x - 1} d x = 9 ln ∣ x - 1 ∣

\int \frac{12}{( x - 1 ) ^{2}} d x = - \frac{12}{x - 1}

\int \frac{- 9 x + 4}{x ^{2} + 1} d x = - \frac{9}{2} ln x^{2} + 1 + 4 arctan (x)

= 9 ln ∣ x - 1 ∣ - (- \frac{12}{x - 1}) - \frac{9}{2} ln x^{2} + 1 + 4 arctan (x)

Simplify

= 9 ln ∣ x - 1 ∣ + \frac{12}{x - 1} - \frac{9}{2} ln x^{2} + 1 + 4 arctan (x)

Add a constant to the solution

= 9 ln ∣ x - 1 ∣ + \frac{12}{x - 1} - \frac{9}{2} ln x^{2} + 1 + 4 arctan (x) + C

\frac{\partial}{\partial x} (41 - x^{2} - y^{2})

\frac{\partial}{\partial x} (sin (2 y))

x \to 1 lim (∣ x ∣)

\int_{0}^{7} t e^{- t} d t

What is the integral of (x^2-8x-17)/((x-1)^2(x^2+1)) ?
The integral of (x^2-8x-17)/((x-1)^2(x^2+1)) is 9ln|x-1|+(12)/(x-1)-9/2 ln|x^2+1|+4arctan(x)+C