What is the integral of (4x^2)/((x-8)^2(x+8)) ?

The integral of (4x^2)/((x-8)^2(x+8)) is 3ln|x-8|-(16)/(x-8)+ln|x+8|+C

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integral of (4x^2)/((x-8)^2(x+8))

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

3 ln ∣ x - 8 ∣ - \frac{16}{x - 8} + ln ∣ x + 8 ∣ + C

Solution steps

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

Take the constant out:

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

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

Take the partial fraction of

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

= 4 \cdot \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

= 4 (\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 ∣

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

Simplify

4 (\frac{3}{4} ln ∣ x - 8 ∣ - \frac{4}{x - 8} + \frac{1}{4} ln ∣ x + 8 ∣) : 3 ln ∣ x - 8 ∣ - \frac{16}{x - 8} + ln ∣ x + 8 ∣

= 3 ln ∣ x - 8 ∣ - \frac{16}{x - 8} + ln ∣ x + 8 ∣

Add a constant to the solution

= 3 ln ∣ x - 8 ∣ - \frac{16}{x - 8} + ln ∣ x + 8 ∣ + C

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What is the integral of (4x^2)/((x-8)^2(x+8)) ?
The integral of (4x^2)/((x-8)^2(x+8)) is 3ln|x-8|-(16)/(x-8)+ln|x+8|+C