What is the integral of (32x^2)/((x-12)(x+4)^2) ?

The integral of (32x^2)/((x-12)(x+4)^2) is 32(9/16 ln|x-12|+7/16 ln|x+4|+1/(x+4))+C

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integral of (32x^2)/((x-12)(x+4)^2)

\int \frac{32 x ^{2}}{( x - 12 ) ( x + 4 ) ^{2}} d x

32 (\frac{9}{16} ln ∣ x - 12 ∣ + \frac{7}{16} ln ∣ x + 4 ∣ + \frac{1}{x + 4}) + C

Solution steps

\int \frac{32 x ^{2}}{( x - 12 ) ( x + 4 ) ^{2}} d x

Take the constant out:

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

= 32 \cdot \int \frac{x ^{2}}{( x - 12 ) ( x + 4 ) ^{2}} d x

Take the partial fraction of

\frac{x ^{2}}{( x - 12 ) ( x + 4 ) ^{2}} : \frac{9}{16 ( x - 12 )} + \frac{7}{16 ( x + 4 )} - \frac{1}{( x + 4 ) ^{2}}

= 32 \cdot \int \frac{9}{16 ( x - 12 )} + \frac{7}{16 ( x + 4 )} - \frac{1}{( 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

= 32 (\int \frac{9}{16 ( x - 12 )} d x + \int \frac{7}{16 ( x + 4 )} d x - \int \frac{1}{( x + 4 ) ^{2}} d x)

\int \frac{9}{16 ( x - 12 )} d x = \frac{9}{16} ln ∣ x - 12 ∣

\int \frac{7}{16 ( x + 4 )} d x = \frac{7}{16} ln ∣ x + 4 ∣

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

= 32 (\frac{9}{16} ln ∣ x - 12 ∣ + \frac{7}{16} ln ∣ x + 4 ∣ - (- \frac{1}{x + 4}))

Simplify

= 32 (\frac{9}{16} ln ∣ x - 12 ∣ + \frac{7}{16} ln ∣ x + 4 ∣ + \frac{1}{x + 4})

Add a constant to the solution

= 32 (\frac{9}{16} ln ∣ x - 12 ∣ + \frac{7}{16} ln ∣ x + 4 ∣ + \frac{1}{x + 4}) + C

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What is the integral of (32x^2)/((x-12)(x+4)^2) ?
The integral of (32x^2)/((x-12)(x+4)^2) is 32(9/16 ln|x-12|+7/16 ln|x+4|+1/(x+4))+C