What is the integral of (25)/((x-1)(x^2+4)) ?

The integral of (25)/((x-1)(x^2+4)) is 25(1/5 ln|x-1|+1/5 (-1/2 ln|x^2+4|-1/2 arctan(x/2)))+C

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integral of (25)/((x-1)(x^2+4))

\int \frac{25}{( x - 1 ) ( x ^{2} + 4 )} d x

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

Solution steps

\int \frac{25}{( x - 1 ) ( x ^{2} + 4 )} d x

Take the constant out:

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

= 25 \cdot \int \frac{1}{( x - 1 ) ( x ^{2} + 4 )} d x

Take the partial fraction of

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

= 25 \cdot \int \frac{1}{5 ( x - 1 )} + \frac{- x - 1}{5 ( x ^{2} + 4 )} 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

= 25 (\int \frac{1}{5 ( x - 1 )} d x + \int \frac{- x - 1}{5 ( x ^{2} + 4 )} d x)

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

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

= 25 (\frac{1}{5} ln ∣ x - 1 ∣ + \frac{1}{5} (- \frac{1}{2} ln x^{2} + 4 - \frac{1}{2} arctan (\frac{x}{2})))

Add a constant to the solution

= 25 (\frac{1}{5} ln ∣ x - 1 ∣ + \frac{1}{5} (- \frac{1}{2} ln x^{2} + 4 - \frac{1}{2} arctan (\frac{x}{2}))) + C

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What is the integral of (25)/((x-1)(x^2+4)) ?
The integral of (25)/((x-1)(x^2+4)) is 25(1/5 ln|x-1|+1/5 (-1/2 ln|x^2+4|-1/2 arctan(x/2)))+C