What is the integral of (x^2-18x-1)/((x-1)^2(x^2+1)) ?

The integral of (x^2-18x-1)/((x-1)^2(x^2+1)) is ln|x-1|+9/(x-1)-1/2 ln|x^2+1|+9arctan(x)+C

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integral of (x^2-18x-1)/((x-1)^2(x^2+1))

\int \frac{x ^{2} - 18 x - 1}{( x - 1 ) ^{2} ( x ^{2} + 1 )} d x

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

Solution steps

\int \frac{x ^{2} - 18 x - 1}{( x - 1 ) ^{2} ( x ^{2} + 1 )} d x

Take the partial fraction of

\frac{x ^{2} - 18 x - 1}{( x - 1 ) ^{2} ( x ^{2} + 1 )} : \frac{1}{x - 1} - \frac{9}{( x - 1 ) ^{2}} + \frac{- x + 9}{x ^{2} + 1}

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

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

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

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

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

Simplify

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

Add a constant to the solution

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

(x + y)^{2} d x + (2 x y + x^{2} - 1) d y = 0, y (1) = 1

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

\frac{\partial}{\partial y} (5 x (4 + y)^{- 1})

5 y^{^{''}} - 6 y^{^{'}} + 5 y = 0

f (x) = \frac{16}{x}

What is the integral of (x^2-18x-1)/((x-1)^2(x^2+1)) ?
The integral of (x^2-18x-1)/((x-1)^2(x^2+1)) is ln|x-1|+9/(x-1)-1/2 ln|x^2+1|+9arctan(x)+C