What is the integral of (x^2-12x-13)/((x-1)^2(x^2+1)) ?

The integral of (x^2-12x-13)/((x-1)^2(x^2+1)) is 7ln|x-1|+(12)/(x-1)-7/2 ln|x^2+1|+6arctan(x)+C

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integral of (x^2-12x-13)/((x-1)^2(x^2+1))

\int \frac{x ^{2} - 12 x - 13}{( x - 1 ) ^{2} ( x ^{2} + 1 )} d x

7 ln ∣ x - 1 ∣ + \frac{12}{x - 1} - \frac{7}{2} ln x^{2} + 1 + 6 arctan (x) + C

Solution steps

\int \frac{x ^{2} - 12 x - 13}{( x - 1 ) ^{2} ( x ^{2} + 1 )} d x

Take the partial fraction of

\frac{x ^{2} - 12 x - 13}{( x - 1 ) ^{2} ( x ^{2} + 1 )} : \frac{7}{x - 1} - \frac{12}{( x - 1 ) ^{2}} + \frac{- 7 x + 6}{x ^{2} + 1}

= \int \frac{7}{x - 1} - \frac{12}{( x - 1 ) ^{2}} + \frac{- 7 x + 6}{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{12}{( x - 1 ) ^{2}} d x + \int \frac{- 7 x + 6}{x ^{2} + 1} d x

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

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

\int \frac{- 7 x + 6}{x ^{2} + 1} d x = - \frac{7}{2} ln x^{2} + 1 + 6 arctan (x)

= 7 ln ∣ x - 1 ∣ - (- \frac{12}{x - 1}) - \frac{7}{2} ln x^{2} + 1 + 6 arctan (x)

Simplify

= 7 ln ∣ x - 1 ∣ + \frac{12}{x - 1} - \frac{7}{2} ln x^{2} + 1 + 6 arctan (x)

Add a constant to the solution

= 7 ln ∣ x - 1 ∣ + \frac{12}{x - 1} - \frac{7}{2} ln x^{2} + 1 + 6 arctan (x) + C

(2 t - 1)^{3}

\int - 7 x e^{x^{2}} d x

\int x^{2} 2 x^{3} + 1 d x

2 y^{^{'}} + 4 y = - 8 y^{3}

What is the integral of (x^2-12x-13)/((x-1)^2(x^2+1)) ?
The integral of (x^2-12x-13)/((x-1)^2(x^2+1)) is 7ln|x-1|+(12)/(x-1)-7/2 ln|x^2+1|+6arctan(x)+C