What is the integral of 1/((x-8)(x^2+25)) ?

The integral of 1/((x-8)(x^2+25)) is 1/89 ln|x-8|+1/89 (-1/2 ln|x^2+25|-8/5 arctan(x/5))+C

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integral of 1/((x-8)(x^2+25))

\int \frac{1}{( x - 8 ) ( x ^{2} + 25 )} d x

\frac{1}{89} ln ∣ x - 8 ∣ + \frac{1}{89} (- \frac{1}{2} ln x^{2} + 25 - \frac{8}{5} arctan (\frac{x}{5})) + C

Solution steps

\int \frac{1}{( x - 8 ) ( x ^{2} + 25 )} d x

Take the partial fraction of

\frac{1}{( x - 8 ) ( x ^{2} + 25 )} : \frac{1}{89 ( x - 8 )} + \frac{- x - 8}{89 ( x ^{2} + 25 )}

= \int \frac{1}{89 ( x - 8 )} + \frac{- x - 8}{89 ( x ^{2} + 25 )} 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}{89 ( x - 8 )} d x + \int \frac{- x - 8}{89 ( x ^{2} + 25 )} d x

\int \frac{1}{89 ( x - 8 )} d x = \frac{1}{89} ln ∣ x - 8 ∣

\int \frac{- x - 8}{89 ( x ^{2} + 25 )} d x = \frac{1}{89} (- \frac{1}{2} ln x^{2} + 25 - \frac{8}{5} arctan (\frac{x}{5}))

= \frac{1}{89} ln ∣ x - 8 ∣ + \frac{1}{89} (- \frac{1}{2} ln x^{2} + 25 - \frac{8}{5} arctan (\frac{x}{5}))

Add a constant to the solution

= \frac{1}{89} ln ∣ x - 8 ∣ + \frac{1}{89} (- \frac{1}{2} ln x^{2} + 25 - \frac{8}{5} arctan (\frac{x}{5})) + C

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

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

y = 6 sin (π x - y), (1, 0)

\int 2 cos (2 x) - 1 d x

cos^{2} (2 t)

What is the integral of 1/((x-8)(x^2+25)) ?
The integral of 1/((x-8)(x^2+25)) is 1/89 ln|x-8|+1/89 (-1/2 ln|x^2+25|-8/5 arctan(x/5))+C