What is the integral of (82)/((x-1)(x^2+81)) ?

The integral of (82)/((x-1)(x^2+81)) is ln|x-1|-1/2 ln|x^2+81|-1/9 arctan(x/9)+C

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integral of (82)/((x-1)(x^2+81))

\int \frac{82}{( x - 1 ) ( x ^{2} + 81 )} d x

ln ∣ x - 1 ∣ - \frac{1}{2} ln x^{2} + 81 - \frac{1}{9} arctan (\frac{x}{9}) + C

Solution steps

\int \frac{82}{( x - 1 ) ( x ^{2} + 81 )} d x

Take the constant out:

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

= 82 \cdot \int \frac{1}{( x - 1 ) ( x ^{2} + 81 )} d x

Take the partial fraction of

\frac{1}{( x - 1 ) ( x ^{2} + 81 )} : \frac{1}{82 ( x - 1 )} + \frac{- x - 1}{82 ( x ^{2} + 81 )}

= 82 \cdot \int \frac{1}{82 ( x - 1 )} + \frac{- x - 1}{82 ( x ^{2} + 81 )} 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

= 82 (\int \frac{1}{82 ( x - 1 )} d x + \int \frac{- x - 1}{82 ( x ^{2} + 81 )} d x)

\int \frac{1}{82 ( x - 1 )} d x = \frac{1}{82} ln ∣ x - 1 ∣

\int \frac{- x - 1}{82 ( x ^{2} + 81 )} d x = \frac{1}{82} (- \frac{1}{2} ln x^{2} + 81 - \frac{1}{9} arctan (\frac{x}{9}))

= 82 (\frac{1}{82} ln ∣ x - 1 ∣ + \frac{1}{82} (- \frac{1}{2} ln x^{2} + 81 - \frac{1}{9} arctan (\frac{x}{9})))

Simplify

82 (\frac{1}{82} ln ∣ x - 1 ∣ + \frac{1}{82} (- \frac{1}{2} ln x^{2} + 81 - \frac{1}{9} arctan (\frac{x}{9}))) : ln ∣ x - 1 ∣ - \frac{1}{2} ln x^{2} + 81 - \frac{1}{9} arctan (\frac{x}{9})

= ln ∣ x - 1 ∣ - \frac{1}{2} ln x^{2} + 81 - \frac{1}{9} arctan (\frac{x}{9})

Add a constant to the solution

= ln ∣ x - 1 ∣ - \frac{1}{2} ln x^{2} + 81 - \frac{1}{9} arctan (\frac{x}{9}) + C

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What is the integral of (82)/((x-1)(x^2+81)) ?
The integral of (82)/((x-1)(x^2+81)) is ln|x-1|-1/2 ln|x^2+81|-1/9 arctan(x/9)+C