What is the integral of (32x^2)/((x-9)(x+3)^2) ?

The integral of (32x^2)/((x-9)(x+3)^2) is 32(9/16 ln|x-9|+7/16 ln|x+3|+3/(4(x+3)))+C

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integral of (32x^2)/((x-9)(x+3)^2)

\int \frac{32 x ^{2}}{( x - 9 ) ( x + 3 ) ^{2}} d x

32 (\frac{9}{16} ln ∣ x - 9 ∣ + \frac{7}{16} ln ∣ x + 3 ∣ + \frac{3}{4 ( x + 3 )}) + C

Solution steps

\int \frac{32 x ^{2}}{( x - 9 ) ( x + 3 ) ^{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 - 9 ) ( x + 3 ) ^{2}} d x

Take the partial fraction of

\frac{x ^{2}}{( x - 9 ) ( x + 3 ) ^{2}} : \frac{9}{16 ( x - 9 )} + \frac{7}{16 ( x + 3 )} - \frac{3}{4 ( x + 3 ) ^{2}}

= 32 \cdot \int \frac{9}{16 ( x - 9 )} + \frac{7}{16 ( x + 3 )} - \frac{3}{4 ( x + 3 ) ^{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 - 9 )} d x + \int \frac{7}{16 ( x + 3 )} d x - \int \frac{3}{4 ( x + 3 ) ^{2}} d x)

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

\int \frac{7}{16 ( x + 3 )} d x = \frac{7}{16} ln ∣ x + 3 ∣

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

= 32 (\frac{9}{16} ln ∣ x - 9 ∣ + \frac{7}{16} ln ∣ x + 3 ∣ - (- \frac{3}{4 ( x + 3 )}))

Simplify

= 32 (\frac{9}{16} ln ∣ x - 9 ∣ + \frac{7}{16} ln ∣ x + 3 ∣ + \frac{3}{4 ( x + 3 )})

Add a constant to the solution

= 32 (\frac{9}{16} ln ∣ x - 9 ∣ + \frac{7}{16} ln ∣ x + 3 ∣ + \frac{3}{4 ( x + 3 )}) + C

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\frac{\partial}{\partial x} (csc (2 x))

\frac{d}{d x} (x - 4 x)

y^{^{'}} (x) = \frac{6 x ^{2}}{6 y ^{2} - 11}, y (1) = 0

\int 16 + x^{2} d x

What is the integral of (32x^2)/((x-9)(x+3)^2) ?
The integral of (32x^2)/((x-9)(x+3)^2) is 32(9/16 ln|x-9|+7/16 ln|x+3|+3/(4(x+3)))+C