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

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

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

\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{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 )}

= \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

= \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}))

= \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

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