What is the integral of ((17))/((x-1)(x^2+16)) ?

The integral of ((17))/((x-1)(x^2+16)) is ln|x-1|-1/2 ln|x^2+16|-1/4 arctan(x/4)+C

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integral of ((17))/((x-1)(x^2+16))

\int \frac{( 17 )}{( x - 1 ) ( x ^{2} + 16 )} d x

ln ∣ x - 1 ∣ - \frac{1}{2} ln x^{2} + 16 - \frac{1}{4} arctan (\frac{x}{4}) + C

Solution steps

\int \frac{17}{( x - 1 ) ( x ^{2} + 16 )} d x

Take the constant out:

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

= 17 \cdot \int \frac{1}{( x - 1 ) ( x ^{2} + 16 )} d x

Take the partial fraction of

\frac{1}{( x - 1 ) ( x ^{2} + 16 )} : \frac{1}{17 ( x - 1 )} + \frac{- x - 1}{17 ( x ^{2} + 16 )}

= 17 \cdot \int \frac{1}{17 ( x - 1 )} + \frac{- x - 1}{17 ( x ^{2} + 16 )} 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

= 17 (\int \frac{1}{17 ( x - 1 )} d x + \int \frac{- x - 1}{17 ( x ^{2} + 16 )} d x)

\int \frac{1}{17 ( x - 1 )} d x = \frac{1}{17} ln ∣ x - 1 ∣

\int \frac{- x - 1}{17 ( x ^{2} + 16 )} d x = \frac{1}{17} (- \frac{1}{2} ln x^{2} + 16 - \frac{1}{4} arctan (\frac{x}{4}))

= 17 (\frac{1}{17} ln ∣ x - 1 ∣ + \frac{1}{17} (- \frac{1}{2} ln x^{2} + 16 - \frac{1}{4} arctan (\frac{x}{4})))

Simplify

17 (\frac{1}{17} ln ∣ x - 1 ∣ + \frac{1}{17} (- \frac{1}{2} ln x^{2} + 16 - \frac{1}{4} arctan (\frac{x}{4}))) : ln ∣ x - 1 ∣ - \frac{1}{2} ln x^{2} + 16 - \frac{1}{4} arctan (\frac{x}{4})

= ln ∣ x - 1 ∣ - \frac{1}{2} ln x^{2} + 16 - \frac{1}{4} arctan (\frac{x}{4})

Add a constant to the solution

= ln ∣ x - 1 ∣ - \frac{1}{2} ln x^{2} + 16 - \frac{1}{4} arctan (\frac{x}{4}) + C

What is the integral of ((17))/((x-1)(x^2+16)) ?
The integral of ((17))/((x-1)(x^2+16)) is ln|x-1|-1/2 ln|x^2+16|-1/4 arctan(x/4)+C