What is the integral of (16x^2)/((x-24)(x+8)^2) ?

The integral of (16x^2)/((x-24)(x+8)^2) is 9ln|x-24|+7ln|x+8|+(32)/(x+8)+C

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integral of (16x^2)/((x-24)(x+8)^2)

\int \frac{16 x ^{2}}{( x - 24 ) ( x + 8 ) ^{2}} d x

9 ln ∣ x - 24 ∣ + 7 ln ∣ x + 8 ∣ + \frac{32}{x + 8} + C

Solution steps

\int \frac{16 x ^{2}}{( x - 24 ) ( x + 8 ) ^{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 - 24 ) ( x + 8 ) ^{2}} d x

Take the partial fraction of

\frac{x ^{2}}{( x - 24 ) ( x + 8 ) ^{2}} : \frac{9}{16 ( x - 24 )} + \frac{7}{16 ( x + 8 )} - \frac{2}{( x + 8 ) ^{2}}

= 16 \cdot \int \frac{9}{16 ( x - 24 )} + \frac{7}{16 ( x + 8 )} - \frac{2}{( x + 8 ) ^{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 - 24 )} d x + \int \frac{7}{16 ( x + 8 )} d x - \int \frac{2}{( x + 8 ) ^{2}} d x)

\int \frac{9}{16 ( x - 24 )} d x = \frac{9}{16} ln ∣ x - 24 ∣

\int \frac{7}{16 ( x + 8 )} d x = \frac{7}{16} ln ∣ x + 8 ∣

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

= 16 (\frac{9}{16} ln ∣ x - 24 ∣ + \frac{7}{16} ln ∣ x + 8 ∣ - (- \frac{2}{x + 8}))

Simplify

16 (\frac{9}{16} ln ∣ x - 24 ∣ + \frac{7}{16} ln ∣ x + 8 ∣ - (- \frac{2}{x + 8})) : 9 ln ∣ x - 24 ∣ + 7 ln ∣ x + 8 ∣ + \frac{32}{x + 8}

= 9 ln ∣ x - 24 ∣ + 7 ln ∣ x + 8 ∣ + \frac{32}{x + 8}

Add a constant to the solution

= 9 ln ∣ x - 24 ∣ + 7 ln ∣ x + 8 ∣ + \frac{32}{x + 8} + C

\int_{0}^{1} \frac{2}{1 - x ^{2}} d x

\frac{\partial}{\partial y} (\frac{x + 1}{y - 1})

\int_{0}^{1} \frac{1}{x ( 1 + x )} d x

\int y y + 1 d y

y = 5 x - x^{2}

What is the integral of (16x^2)/((x-24)(x+8)^2) ?
The integral of (16x^2)/((x-24)(x+8)^2) is 9ln|x-24|+7ln|x+8|+(32)/(x+8)+C