What is the integral of (16x^2)/((x-12)(x+4)^2) ?

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

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integral of (16x^2)/((x-12)(x+4)^2)

\int \frac{16 x ^{2}}{( x - 12 ) ( x + 4 ) ^{2}} d x

9 ln ∣ x - 12 ∣ + 7 ln ∣ x + 4 ∣ + \frac{16}{x + 4} + C

Solution steps

\int \frac{16 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

= 16 \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}}

= 16 \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

= 16 (\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}

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

Simplify

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

= 9 ln ∣ x - 12 ∣ + 7 ln ∣ x + 4 ∣ + \frac{16}{x + 4}

Add a constant to the solution

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

\int (\frac{1}{8}) y_{1} e^{- \frac{y _{1} + y _{2}}{2}} d y_{1}

f (x) = 3 sin (x) - 4 cos (x)

y^{^{'}} = 4 x y

\int_{- 2}^{1} 1 + 25 d x

x \to 27 lim (\frac{3 - 3 x}{27 - x})

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