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

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

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

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

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

Solution steps

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

Take the constant out:

\int a \cdot f (x) d x = a \cdot \int f (x) d x

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

Take the partial fraction of

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

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

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

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

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

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

Simplify

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

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

Add a constant to the solution

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

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