What is the integral of (16x^2)/((x-21)(x+7)^2) ?

The integral of (16x^2)/((x-21)(x+7)^2) is 9ln|x-21|+7ln|x+7|+(28)/(x+7)+C

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integral of (16x^2)/((x-21)(x+7)^2)

\int \frac{16 x ^{2}}{( x - 21 ) ( x + 7 ) ^{2}} d x

9 ln ∣ x - 21 ∣ + 7 ln ∣ x + 7 ∣ + \frac{28}{x + 7} + C

Solution steps

\int \frac{16 x ^{2}}{( x - 21 ) ( x + 7 ) ^{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 - 21 ) ( x + 7 ) ^{2}} d x

Take the partial fraction of

\frac{x ^{2}}{( x - 21 ) ( x + 7 ) ^{2}} : \frac{9}{16 ( x - 21 )} + \frac{7}{16 ( x + 7 )} - \frac{7}{4 ( x + 7 ) ^{2}}

= 16 \cdot \int \frac{9}{16 ( x - 21 )} + \frac{7}{16 ( x + 7 )} - \frac{7}{4 ( x + 7 ) ^{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 - 21 )} d x + \int \frac{7}{16 ( x + 7 )} d x - \int \frac{7}{4 ( x + 7 ) ^{2}} d x)

\int \frac{9}{16 ( x - 21 )} d x = \frac{9}{16} ln ∣ x - 21 ∣

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

\int \frac{7}{4 ( x + 7 ) ^{2}} d x = - \frac{7}{4 ( x + 7 )}

= 16 (\frac{9}{16} ln ∣ x - 21 ∣ + \frac{7}{16} ln ∣ x + 7 ∣ - (- \frac{7}{4 ( x + 7 )}))

Simplify

16 (\frac{9}{16} ln ∣ x - 21 ∣ + \frac{7}{16} ln ∣ x + 7 ∣ - (- \frac{7}{4 ( x + 7 )})) : 9 ln ∣ x - 21 ∣ + 7 ln ∣ x + 7 ∣ + \frac{28}{x + 7}

= 9 ln ∣ x - 21 ∣ + 7 ln ∣ x + 7 ∣ + \frac{28}{x + 7}

Add a constant to the solution

= 9 ln ∣ x - 21 ∣ + 7 ln ∣ x + 7 ∣ + \frac{28}{x + 7} + C

\frac{\partial}{\partial x} (\frac{2 x}{x ^{2} + y - 5})

\frac{d y}{d x} = b \cdot (1 - y) \cdot y

\int_{0}^{\frac{π}{3}} sin^{2} (x) d x

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

y^{2} - 5 x = 4, x - y = 2

What is the integral of (16x^2)/((x-21)(x+7)^2) ?
The integral of (16x^2)/((x-21)(x+7)^2) is 9ln|x-21|+7ln|x+7|+(28)/(x+7)+C