What is the integral of (x+5)/((x-1)^2(x^2+1)) ?

The integral of (x+5)/((x-1)^2(x^2+1)) is -1/2 arctan(x)-5/2 ln|x-1|+5/4 ln|x^2+1|-3/(x-1)+C

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integral of (x+5)/((x-1)^2(x^2+1))

\int \frac{x + 5}{( x - 1 ) ^{2} ( x ^{2} + 1 )} d x

- \frac{1}{2} arctan (x) - \frac{5}{2} ln ∣ x - 1 ∣ + \frac{5}{4} ln x^{2} + 1 - \frac{3}{x - 1} + C

Solution steps

\int \frac{x + 5}{( x - 1 ) ^{2} ( x ^{2} + 1 )} d x

Expand

\frac{x + 5}{( x - 1 ) ^{2} ( x ^{2} + 1 )} : \frac{x}{( x - 1 ) ^{2} ( x ^{2} + 1 )} + \frac{5}{( x - 1 ) ^{2} ( x ^{2} + 1 )}

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{x}{( x - 1 ) ^{2} ( x ^{2} + 1 )} d x + \int \frac{5}{( x - 1 ) ^{2} ( x ^{2} + 1 )} d x

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

\int \frac{5}{( x - 1 ) ^{2} ( x ^{2} + 1 )} d x = 5 (- \frac{1}{2} ln ∣ x - 1 ∣ - \frac{1}{2 ( x - 1 )} + \frac{1}{4} ln x^{2} + 1)

= - \frac{1}{2 ( x - 1 )} - \frac{1}{2} arctan (x) + 5 (- \frac{1}{2} ln ∣ x - 1 ∣ - \frac{1}{2 ( x - 1 )} + \frac{1}{4} ln x^{2} + 1)

Simplify

- \frac{1}{2 ( x - 1 )} - \frac{1}{2} arctan (x) + 5 (- \frac{1}{2} ln ∣ x - 1 ∣ - \frac{1}{2 ( x - 1 )} + \frac{1}{4} ln x^{2} + 1) : - \frac{1}{2} arctan (x) - \frac{5}{2} ln ∣ x - 1 ∣ + \frac{5}{4} ln x^{2} + 1 - \frac{3}{x - 1}

= - \frac{1}{2} arctan (x) - \frac{5}{2} ln ∣ x - 1 ∣ + \frac{5}{4} ln x^{2} + 1 - \frac{3}{x - 1}

Add a constant to the solution

= - \frac{1}{2} arctan (x) - \frac{5}{2} ln ∣ x - 1 ∣ + \frac{5}{4} ln x^{2} + 1 - \frac{3}{x - 1} + C

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What is the integral of (x+5)/((x-1)^2(x^2+1)) ?
The integral of (x+5)/((x-1)^2(x^2+1)) is -1/2 arctan(x)-5/2 ln|x-1|+5/4 ln|x^2+1|-3/(x-1)+C