What is the integral of (x^2-6x-17)/((x-1)^2(x^2+1)) ?

The integral of (x^2-6x-17)/((x-1)^2(x^2+1)) is 9ln|x-1|+(11)/(x-1)-9/2 ln|x^2+1|+3arctan(x)+C

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integral of (x^2-6x-17)/((x-1)^2(x^2+1))

\int \frac{x ^{2} - 6 x - 17}{( x - 1 ) ^{2} ( x ^{2} + 1 )} d x

9 ln ∣ x - 1 ∣ + \frac{11}{x - 1} - \frac{9}{2} ln x^{2} + 1 + 3 arctan (x) + C

Solution steps

\int \frac{x ^{2} - 6 x - 17}{( x - 1 ) ^{2} ( x ^{2} + 1 )} d x

Take the partial fraction of

\frac{x ^{2} - 6 x - 17}{( x - 1 ) ^{2} ( x ^{2} + 1 )} : \frac{9}{x - 1} - \frac{11}{( x - 1 ) ^{2}} + \frac{- 9 x + 3}{x ^{2} + 1}

= \int \frac{9}{x - 1} - \frac{11}{( x - 1 ) ^{2}} + \frac{- 9 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{9}{x - 1} d x - \int \frac{11}{( x - 1 ) ^{2}} d x + \int \frac{- 9 x + 3}{x ^{2} + 1} d x

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

\int \frac{11}{( x - 1 ) ^{2}} d x = - \frac{11}{x - 1}

\int \frac{- 9 x + 3}{x ^{2} + 1} d x = - \frac{9}{2} ln x^{2} + 1 + 3 arctan (x)

= 9 ln ∣ x - 1 ∣ - (- \frac{11}{x - 1}) - \frac{9}{2} ln x^{2} + 1 + 3 arctan (x)

Simplify

= 9 ln ∣ x - 1 ∣ + \frac{11}{x - 1} - \frac{9}{2} ln x^{2} + 1 + 3 arctan (x)

Add a constant to the solution

= 9 ln ∣ x - 1 ∣ + \frac{11}{x - 1} - \frac{9}{2} ln x^{2} + 1 + 3 arctan (x) + C

\int_{0}^{3} (9 - x^{2}) d x

\frac{\partial}{\partial x} (3 x y^{2} + 9 x)

f (x) = x \cdot 2 x + 2

y^{^{'}} = e^{(x + y)}

h (x) = (5 - 5 x^{3})^{3} + \frac{1}{5 - 5 x ^{3}}

What is the integral of (x^2-6x-17)/((x-1)^2(x^2+1)) ?
The integral of (x^2-6x-17)/((x-1)^2(x^2+1)) is 9ln|x-1|+(11)/(x-1)-9/2 ln|x^2+1|+3arctan(x)+C