What is the integral of (16x^2)/((x-9)(x+3)^2) ?

The integral of (16x^2)/((x-9)(x+3)^2) is 9ln|x-9|+7ln|x+3|+(12)/(x+3)+C

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integral of (16x^2)/((x-9)(x+3)^2)

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

9 ln ∣ x - 9 ∣ + 7 ln ∣ x + 3 ∣ + \frac{12}{x + 3} + C

Solution steps

\int \frac{16 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

= 16 \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}}

= 16 \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

= 16 (\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 )}

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

Simplify

16 (\frac{9}{16} ln ∣ x - 9 ∣ + \frac{7}{16} ln ∣ x + 3 ∣ - (- \frac{3}{4 ( x + 3 )})) : 9 ln ∣ x - 9 ∣ + 7 ln ∣ x + 3 ∣ + \frac{12}{x + 3}

= 9 ln ∣ x - 9 ∣ + 7 ln ∣ x + 3 ∣ + \frac{12}{x + 3}

Add a constant to the solution

= 9 ln ∣ x - 9 ∣ + 7 ln ∣ x + 3 ∣ + \frac{12}{x + 3} + C

\frac{d}{d x} ((5 x - 4) (2 x^{3} - x^{2} + 1))

f (x) = \frac{1}{4 + 3 x}, (1, \frac{1}{7})

e^{9 x}

f (x) = 2 x^{2} - 4 x + 1

\int \frac{1}{x ^{\frac{5}{8}} - x ^{\frac{1}{8}}} d x

What is the integral of (16x^2)/((x-9)(x+3)^2) ?
The integral of (16x^2)/((x-9)(x+3)^2) is 9ln|x-9|+7ln|x+3|+(12)/(x+3)+C