What is the integral of (16x^2)/((x-18)(x+6)^2) ?

The integral of (16x^2)/((x-18)(x+6)^2) is 9ln|x-18|+7ln|x+6|+(24)/(x+6)+C

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integral of (16x^2)/((x-18)(x+6)^2)

\int \frac{16 x ^{2}}{( x - 18 ) ( x + 6 ) ^{2}} d x

9 ln ∣ x - 18 ∣ + 7 ln ∣ x + 6 ∣ + \frac{24}{x + 6} + C

Solution steps

\int \frac{16 x ^{2}}{( x - 18 ) ( x + 6 ) ^{2}} d x

Take the constant out:

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

= 16 \cdot \int \frac{x ^{2}}{( x - 18 ) ( x + 6 ) ^{2}} d x

Take the partial fraction of

\frac{x ^{2}}{( x - 18 ) ( x + 6 ) ^{2}} : \frac{9}{16 ( x - 18 )} + \frac{7}{16 ( x + 6 )} - \frac{3}{2 ( x + 6 ) ^{2}}

= 16 \cdot \int \frac{9}{16 ( x - 18 )} + \frac{7}{16 ( x + 6 )} - \frac{3}{2 ( x + 6 ) ^{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

= 16 (\int \frac{9}{16 ( x - 18 )} d x + \int \frac{7}{16 ( x + 6 )} d x - \int \frac{3}{2 ( x + 6 ) ^{2}} d x)

\int \frac{9}{16 ( x - 18 )} d x = \frac{9}{16} ln ∣ x - 18 ∣

\int \frac{7}{16 ( x + 6 )} d x = \frac{7}{16} ln ∣ x + 6 ∣

\int \frac{3}{2 ( x + 6 ) ^{2}} d x = - \frac{3}{2 ( x + 6 )}

= 16 (\frac{9}{16} ln ∣ x - 18 ∣ + \frac{7}{16} ln ∣ x + 6 ∣ - (- \frac{3}{2 ( x + 6 )}))

Simplify

16 (\frac{9}{16} ln ∣ x - 18 ∣ + \frac{7}{16} ln ∣ x + 6 ∣ - (- \frac{3}{2 ( x + 6 )})) : 9 ln ∣ x - 18 ∣ + 7 ln ∣ x + 6 ∣ + \frac{24}{x + 6}

= 9 ln ∣ x - 18 ∣ + 7 ln ∣ x + 6 ∣ + \frac{24}{x + 6}

Add a constant to the solution

= 9 ln ∣ x - 18 ∣ + 7 ln ∣ x + 6 ∣ + \frac{24}{x + 6} + C

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What is the integral of (16x^2)/((x-18)(x+6)^2) ?
The integral of (16x^2)/((x-18)(x+6)^2) is 9ln|x-18|+7ln|x+6|+(24)/(x+6)+C