What is the integral of 1/((x^2+24)(4-x)^2) ?

The integral of 1/((x^2+24)(4-x)^2) is

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integral of 1/((x^2+24)(4-x)^2)

\int \frac{1}{( x ^{2} + 24 ) ( 4 - x ) ^{2}} d x

\frac{1}{200} (\frac{1}{2} ln x^{2} + 24 - \frac{1}{2 6} arctan (\frac{x}{2 6})) - \frac{1}{200} ln ∣ x - 4 ∣ - \frac{1}{40 ( x - 4 )} + C

Solution steps

\int \frac{1}{( x ^{2} + 24 ) ( 4 - x ) ^{2}} d x

Take the partial fraction of

\frac{1}{( x ^{2} + 24 ) ( 4 - x ) ^{2}} : \frac{x - 1}{200 ( x ^{2} + 24 )} - \frac{1}{200 ( x - 4 )} + \frac{1}{40 ( x - 4 ) ^{2}}

= \int \frac{x - 1}{200 ( x ^{2} + 24 )} - \frac{1}{200 ( x - 4 )} + \frac{1}{40 ( x - 4 ) ^{2}} 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{x - 1}{200 ( x ^{2} + 24 )} d x - \int \frac{1}{200 ( x - 4 )} d x + \int \frac{1}{40 ( x - 4 ) ^{2}} d x

\int \frac{x - 1}{200 ( x ^{2} + 24 )} d x = \frac{1}{200} (\frac{1}{2} ln x^{2} + 24 - \frac{1}{2 6} arctan (\frac{x}{2 6}))

\int \frac{1}{200 ( x - 4 )} d x = \frac{1}{200} ln ∣ x - 4 ∣

\int \frac{1}{40 ( x - 4 ) ^{2}} d x = - \frac{1}{40 ( x - 4 )}

= \frac{1}{200} (\frac{1}{2} ln x^{2} + 24 - \frac{1}{2 6} arctan (\frac{x}{2 6})) - \frac{1}{200} ln ∣ x - 4 ∣ - \frac{1}{40 ( x - 4 )}

Add a constant to the solution

= \frac{1}{200} (\frac{1}{2} ln x^{2} + 24 - \frac{1}{2 6} arctan (\frac{x}{2 6})) - \frac{1}{200} ln ∣ x - 4 ∣ - \frac{1}{40 ( x - 4 )} + C

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What is the integral of 1/((x^2+24)(4-x)^2) ?
The integral of 1/((x^2+24)(4-x)^2) is