What is the integral of 1/((x-9)(x^2+16)) ?

The integral of 1/((x-9)(x^2+16)) is 1/97 ln|x-9|+1/97 (-1/2 ln|x^2+16|-9/4 arctan(x/4))+C

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integral of 1/((x-9)(x^2+16))

\int \frac{1}{( x - 9 ) ( x ^{2} + 16 )} d x

\frac{1}{97} ln ∣ x - 9 ∣ + \frac{1}{97} (- \frac{1}{2} ln x^{2} + 16 - \frac{9}{4} arctan (\frac{x}{4})) + C

Solution steps

\int \frac{1}{( x - 9 ) ( x ^{2} + 16 )} d x

Take the partial fraction of

\frac{1}{( x - 9 ) ( x ^{2} + 16 )} : \frac{1}{97 ( x - 9 )} + \frac{- x - 9}{97 ( x ^{2} + 16 )}

= \int \frac{1}{97 ( x - 9 )} + \frac{- x - 9}{97 ( 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

= \int \frac{1}{97 ( x - 9 )} d x + \int \frac{- x - 9}{97 ( x ^{2} + 16 )} d x

\int \frac{1}{97 ( x - 9 )} d x = \frac{1}{97} ln ∣ x - 9 ∣

\int \frac{- x - 9}{97 ( x ^{2} + 16 )} d x = \frac{1}{97} (- \frac{1}{2} ln x^{2} + 16 - \frac{9}{4} arctan (\frac{x}{4}))

= \frac{1}{97} ln ∣ x - 9 ∣ + \frac{1}{97} (- \frac{1}{2} ln x^{2} + 16 - \frac{9}{4} arctan (\frac{x}{4}))

Add a constant to the solution

= \frac{1}{97} ln ∣ x - 9 ∣ + \frac{1}{97} (- \frac{1}{2} ln x^{2} + 16 - \frac{9}{4} arctan (\frac{x}{4})) + C

\frac{d}{d x} (\frac{x ^{2} + 5}{x - 2})

(6 - 2 x^{2})^{3}

n = 0 \sum \infty n (\frac{3}{5})^{n}

25 x^{2} y^{^{''}} + 25 x y^{^{'}} + y = 0

x \to \infty lim (x^{x})

What is the integral of 1/((x-9)(x^2+16)) ?
The integral of 1/((x-9)(x^2+16)) is 1/97 ln|x-9|+1/97 (-1/2 ln|x^2+16|-9/4 arctan(x/4))+C