What is the integral of (20x)/((x-1)(x^2+4x+5)) ?

The integral of (20x)/((x-1)(x^2+4x+5)) is 2ln|x-1|-ln|x^2+4x+5|+14arctan(x+2)+C

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integral of (20x)/((x-1)(x^2+4x+5))

\int \frac{20 x}{( x - 1 ) ( x ^{2} + 4 x + 5 )} d x

2 ln ∣ x - 1 ∣ - ln x^{2} + 4 x + 5 + 14 arctan (x + 2) + C

Solution steps

\int \frac{20 x}{( x - 1 ) ( x ^{2} + 4 x + 5 )} d x

Take the constant out:

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

= 20 \cdot \int \frac{x}{( x - 1 ) ( x ^{2} + 4 x + 5 )} d x

Take the partial fraction of

\frac{x}{( x - 1 ) ( x ^{2} + 4 x + 5 )} : \frac{1}{10 ( x - 1 )} + \frac{- x + 5}{10 ( x ^{2} + 4 x + 5 )}

= 20 \cdot \int \frac{1}{10 ( x - 1 )} + \frac{- x + 5}{10 ( x ^{2} + 4 x + 5 )} 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

= 20 (\int \frac{1}{10 ( x - 1 )} d x + \int \frac{- x + 5}{10 ( x ^{2} + 4 x + 5 )} d x)

\int \frac{1}{10 ( x - 1 )} d x = \frac{1}{10} ln ∣ x - 1 ∣

\int \frac{- x + 5}{10 ( x ^{2} + 4 x + 5 )} d x = \frac{1}{20} (- ln x^{2} + 4 x + 5 + 14 arctan (x + 2))

= 20 (\frac{1}{10} ln ∣ x - 1 ∣ + \frac{1}{20} (- ln x^{2} + 4 x + 5 + 14 arctan (x + 2)))

Simplify

20 (\frac{1}{10} ln ∣ x - 1 ∣ + \frac{1}{20} (- ln x^{2} + 4 x + 5 + 14 arctan (x + 2))) : 2 ln ∣ x - 1 ∣ - ln x^{2} + 4 x + 5 + 14 arctan (x + 2)

= 2 ln ∣ x - 1 ∣ - ln x^{2} + 4 x + 5 + 14 arctan (x + 2)

Add a constant to the solution

= 2 ln ∣ x - 1 ∣ - ln x^{2} + 4 x + 5 + 14 arctan (x + 2) + C

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What is the integral of (20x)/((x-1)(x^2+4x+5)) ?
The integral of (20x)/((x-1)(x^2+4x+5)) is 2ln|x-1|-ln|x^2+4x+5|+14arctan(x+2)+C