What is the integral of (x^2-8x-13)/((x-1)^2(x^2+1)) ?

The integral of (x^2-8x-13)/((x-1)^2(x^2+1)) is 7ln|x-1|+(10)/(x-1)-7/2 ln|x^2+1|+4arctan(x)+C

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integral of (x^2-8x-13)/((x-1)^2(x^2+1))

\int \frac{x ^{2} - 8 x - 13}{( x - 1 ) ^{2} ( x ^{2} + 1 )} d x

7 ln ∣ x - 1 ∣ + \frac{10}{x - 1} - \frac{7}{2} ln x^{2} + 1 + 4 arctan (x) + C

Solution steps

\int \frac{x ^{2} - 8 x - 13}{( x - 1 ) ^{2} ( x ^{2} + 1 )} d x

Take the partial fraction of

\frac{x ^{2} - 8 x - 13}{( x - 1 ) ^{2} ( x ^{2} + 1 )} : \frac{7}{x - 1} - \frac{10}{( x - 1 ) ^{2}} + \frac{- 7 x + 4}{x ^{2} + 1}

= \int \frac{7}{x - 1} - \frac{10}{( x - 1 ) ^{2}} + \frac{- 7 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{7}{x - 1} d x - \int \frac{10}{( x - 1 ) ^{2}} d x + \int \frac{- 7 x + 4}{x ^{2} + 1} d x

\int \frac{7}{x - 1} d x = 7 ln ∣ x - 1 ∣

\int \frac{10}{( x - 1 ) ^{2}} d x = - \frac{10}{x - 1}

\int \frac{- 7 x + 4}{x ^{2} + 1} d x = - \frac{7}{2} ln x^{2} + 1 + 4 arctan (x)

= 7 ln ∣ x - 1 ∣ - (- \frac{10}{x - 1}) - \frac{7}{2} ln x^{2} + 1 + 4 arctan (x)

Simplify

= 7 ln ∣ x - 1 ∣ + \frac{10}{x - 1} - \frac{7}{2} ln x^{2} + 1 + 4 arctan (x)

Add a constant to the solution

= 7 ln ∣ x - 1 ∣ + \frac{10}{x - 1} - \frac{7}{2} ln x^{2} + 1 + 4 arctan (x) + C

2 x = y^{2}, x + y = 4, x + y = 12

\frac{d}{d x} (x e^{x a})

x \to \infty lim (- 1^{x} x)

\frac{d}{d x} (4 sin^{5} (x))

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

What is the integral of (x^2-8x-13)/((x-1)^2(x^2+1)) ?
The integral of (x^2-8x-13)/((x-1)^2(x^2+1)) is 7ln|x-1|+(10)/(x-1)-7/2 ln|x^2+1|+4arctan(x)+C