What is the integral of (x^2)/((x-8)(x^2-64)) ?

The integral of (x^2)/((x-8)(x^2-64)) is 3/4 ln|x-8|-4/(x-8)+1/4 ln|x+8|+C

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integral of (x^2)/((x-8)(x^2-64))

\int \frac{x ^{2}}{( x - 8 ) ( x ^{2} - 64 )} d x

\frac{3}{4} ln ∣ x - 8 ∣ - \frac{4}{x - 8} + \frac{1}{4} ln ∣ x + 8 ∣ + C

Solution steps

\int \frac{x ^{2}}{( x - 8 ) ( x ^{2} - 64 )} d x

Take the partial fraction of

\frac{x ^{2}}{( x - 8 ) ( x ^{2} - 64 )} : \frac{3}{4 ( x - 8 )} + \frac{4}{( x - 8 ) ^{2}} + \frac{1}{4 ( x + 8 )}

= \int \frac{3}{4 ( x - 8 )} + \frac{4}{( x - 8 ) ^{2}} + \frac{1}{4 ( x + 8 )} 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{3}{4 ( x - 8 )} d x + \int \frac{4}{( x - 8 ) ^{2}} d x + \int \frac{1}{4 ( x + 8 )} d x

\int \frac{3}{4 ( x - 8 )} d x = \frac{3}{4} ln ∣ x - 8 ∣

\int \frac{4}{( x - 8 ) ^{2}} d x = - \frac{4}{x - 8}

\int \frac{1}{4 ( x + 8 )} d x = \frac{1}{4} ln ∣ x + 8 ∣

= \frac{3}{4} ln ∣ x - 8 ∣ - \frac{4}{x - 8} + \frac{1}{4} ln ∣ x + 8 ∣

Add a constant to the solution

= \frac{3}{4} ln ∣ x - 8 ∣ - \frac{4}{x - 8} + \frac{1}{4} ln ∣ x + 8 ∣ + C

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What is the integral of (x^2)/((x-8)(x^2-64)) ?
The integral of (x^2)/((x-8)(x^2-64)) is 3/4 ln|x-8|-4/(x-8)+1/4 ln|x+8|+C