What is the integral of (-8x^2+16x+8)/((x-1)^2(x^2+1)) ?

The integral of (-8x^2+16x+8)/((x-1)^2(x^2+1)) is -8ln|x-1|-8/(x-1)+4ln|x^2+1|-8arctan(x)+C

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integral of (-8x^2+16x+8)/((x-1)^2(x^2+1))

\int \frac{- 8 x ^{2} + 16 x + 8}{( x - 1 ) ^{2} ( x ^{2} + 1 )} d x

- 8 ln ∣ x - 1 ∣ - \frac{8}{x - 1} + 4 ln x^{2} + 1 - 8 arctan (x) + C

Solution steps

\int \frac{- 8 x ^{2} + 16 x + 8}{( x - 1 ) ^{2} ( x ^{2} + 1 )} d x

Take the partial fraction of

\frac{- 8 x ^{2} + 16 x + 8}{( x - 1 ) ^{2} ( x ^{2} + 1 )} : - \frac{8}{x - 1} + \frac{8}{( x - 1 ) ^{2}} + \frac{8 x - 8}{x ^{2} + 1}

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

\int \frac{8}{x - 1} d x = 8 ln ∣ x - 1 ∣

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

\int \frac{8 x - 8}{x ^{2} + 1} d x = 4 ln x^{2} + 1 - 8 arctan (x)

= - 8 ln ∣ x - 1 ∣ - \frac{8}{x - 1} + 4 ln x^{2} + 1 - 8 arctan (x)

Add a constant to the solution

= - 8 ln ∣ x - 1 ∣ - \frac{8}{x - 1} + 4 ln x^{2} + 1 - 8 arctan (x) + C

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x \to 3 lim (5 x^{3} - 3 x + x - 6)

\frac{\partial}{\partial x} (am x^{b} m y^{1 - b})

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

What is the integral of (-8x^2+16x+8)/((x-1)^2(x^2+1)) ?
The integral of (-8x^2+16x+8)/((x-1)^2(x^2+1)) is -8ln|x-1|-8/(x-1)+4ln|x^2+1|-8arctan(x)+C