What is the integral of (16x^2)/((x-6)(x+2)^2) ?

The integral of (16x^2)/((x-6)(x+2)^2) is 7ln|x+2|+8/(x+2)+9ln|x-6|+C

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integral of (16x^2)/((x-6)(x+2)^2)

\int \frac{16 x ^{2}}{( x - 6 ) ( x + 2 ) ^{2}} d x

7 ln ∣ x + 2 ∣ + \frac{8}{x + 2} + 9 ln ∣ x - 6 ∣ + C

Solution steps

\int \frac{16 x ^{2}}{( x - 6 ) ( x + 2 ) ^{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 - 6 ) ( x + 2 ) ^{2}} d x

Apply u-substitution

: \int \frac{( u - 2 ) ^{2}}{u ^{2} ( u - 8 )} d u

= 16 \cdot \int \frac{( u - 2 ) ^{2}}{u ^{2} ( u - 8 )} d u

Take the partial fraction of

\frac{( u - 2 ) ^{2}}{u ^{2} ( u - 8 )} : \frac{7}{16 u} - \frac{1}{2 u ^{2}} + \frac{9}{16 ( u - 8 )}

= 16 \cdot \int \frac{7}{16 u} - \frac{1}{2 u ^{2}} + \frac{9}{16 ( u - 8 )} d u

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{7}{16 u} d u - \int \frac{1}{2 u ^{2}} d u + \int \frac{9}{16 ( u - 8 )} d u)

\int \frac{7}{16 u} d u = \frac{7}{16} ln ∣ u ∣

\int \frac{1}{2 u ^{2}} d u = - \frac{1}{2 u}

\int \frac{9}{16 ( u - 8 )} d u = \frac{9}{16} ln ∣ u - 8 ∣

= 16 (\frac{7}{16} ln ∣ u ∣ - (- \frac{1}{2 u}) + \frac{9}{16} ln ∣ u - 8 ∣)

Substitute back

u = x + 2

= 16 (\frac{7}{16} ln ∣ x + 2 ∣ - (- \frac{1}{2 ( x + 2 )}) + \frac{9}{16} ln ∣ x + 2 - 8 ∣)

Simplify

16 (\frac{7}{16} ln ∣ x + 2 ∣ - (- \frac{1}{2 ( x + 2 )}) + \frac{9}{16} ln ∣ x + 2 - 8 ∣) : 7 ln ∣ x + 2 ∣ + \frac{8}{x + 2} + 9 ln ∣ x - 6 ∣

= 7 ln ∣ x + 2 ∣ + \frac{8}{x + 2} + 9 ln ∣ x - 6 ∣

Add a constant to the solution

= 7 ln ∣ x + 2 ∣ + \frac{8}{x + 2} + 9 ln ∣ x - 6 ∣ + C

\frac{( k ^{2} x ^{3} - 6 k x + 9 )}{x - 1}

\int \frac{x}{2 x ^{2} + 12 x + 23} d x

x (x + y) y^{^{'}} = y (x - y)

\int sin (u) cos (u) d u

f (x) = cos (x)

What is the integral of (16x^2)/((x-6)(x+2)^2) ?
The integral of (16x^2)/((x-6)(x+2)^2) is 7ln|x+2|+8/(x+2)+9ln|x-6|+C