What is the integral of (37)/((x-1)(x^2+36)) ?

The integral of (37)/((x-1)(x^2+36)) is ln|x-1|-1/2 ln|x^2+36|-1/6 arctan(x/6)+C

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integral of (37)/((x-1)(x^2+36))

\int \frac{37}{( x - 1 ) ( x ^{2} + 36 )} d x

ln ∣ x - 1 ∣ - \frac{1}{2} ln x^{2} + 36 - \frac{1}{6} arctan (\frac{x}{6}) + C

Solution steps

\int \frac{37}{( x - 1 ) ( x ^{2} + 36 )} d x

Take the constant out:

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

= 37 \cdot \int \frac{1}{( x - 1 ) ( x ^{2} + 36 )} d x

Take the partial fraction of

\frac{1}{( x - 1 ) ( x ^{2} + 36 )} : \frac{1}{37 ( x - 1 )} + \frac{- x - 1}{37 ( x ^{2} + 36 )}

= 37 \cdot \int \frac{1}{37 ( x - 1 )} + \frac{- x - 1}{37 ( x ^{2} + 36 )} 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

= 37 (\int \frac{1}{37 ( x - 1 )} d x + \int \frac{- x - 1}{37 ( x ^{2} + 36 )} d x)

\int \frac{1}{37 ( x - 1 )} d x = \frac{1}{37} ln ∣ x - 1 ∣

\int \frac{- x - 1}{37 ( x ^{2} + 36 )} d x = \frac{1}{37} (- \frac{1}{2} ln x^{2} + 36 - \frac{1}{6} arctan (\frac{x}{6}))

= 37 (\frac{1}{37} ln ∣ x - 1 ∣ + \frac{1}{37} (- \frac{1}{2} ln x^{2} + 36 - \frac{1}{6} arctan (\frac{x}{6})))

Simplify

37 (\frac{1}{37} ln ∣ x - 1 ∣ + \frac{1}{37} (- \frac{1}{2} ln x^{2} + 36 - \frac{1}{6} arctan (\frac{x}{6}))) : ln ∣ x - 1 ∣ - \frac{1}{2} ln x^{2} + 36 - \frac{1}{6} arctan (\frac{x}{6})

= ln ∣ x - 1 ∣ - \frac{1}{2} ln x^{2} + 36 - \frac{1}{6} arctan (\frac{x}{6})

Add a constant to the solution

= ln ∣ x - 1 ∣ - \frac{1}{2} ln x^{2} + 36 - \frac{1}{6} arctan (\frac{x}{6}) + C

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What is the integral of (37)/((x-1)(x^2+36)) ?
The integral of (37)/((x-1)(x^2+36)) is ln|x-1|-1/2 ln|x^2+36|-1/6 arctan(x/6)+C