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

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

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

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

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

Solution steps

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

Take the partial fraction of

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

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

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

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

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

Simplify

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

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

Add a constant to the solution

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

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