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

The integral of (x^2)/((x+2)^2(x+4)^2) is -2ln|x+2|-1/(x+2)+2ln|x+4|-4/(x+4)+C

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integral of (x^2)/((x+2)^2(x+4)^2)

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

- 2 ln ∣ x + 2 ∣ - \frac{1}{x + 2} + 2 ln ∣ x + 4 ∣ - \frac{4}{x + 4} + C

Solution steps

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

Take the partial fraction of

\frac{x ^{2}}{( x + 2 ) ^{2} ( x + 4 ) ^{2}} : - \frac{2}{x + 2} + \frac{1}{( x + 2 ) ^{2}} + \frac{2}{x + 4} + \frac{4}{( x + 4 ) ^{2}}

= \int - \frac{2}{x + 2} + \frac{1}{( x + 2 ) ^{2}} + \frac{2}{x + 4} + \frac{4}{( 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

= - \int \frac{2}{x + 2} d x + \int \frac{1}{( x + 2 ) ^{2}} d x + \int \frac{2}{x + 4} d x + \int \frac{4}{( x + 4 ) ^{2}} d x

\int \frac{2}{x + 2} d x = 2 ln ∣ x + 2 ∣

\int \frac{1}{( x + 2 ) ^{2}} d x = - \frac{1}{x + 2}

\int \frac{2}{x + 4} d x = 2 ln ∣ x + 4 ∣

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

= - 2 ln ∣ x + 2 ∣ - \frac{1}{x + 2} + 2 ln ∣ x + 4 ∣ - \frac{4}{x + 4}

Add a constant to the solution

= - 2 ln ∣ x + 2 ∣ - \frac{1}{x + 2} + 2 ln ∣ x + 4 ∣ - \frac{4}{x + 4} + C

\int \frac{2 x}{x ^{2} - 2} d x

\frac{d}{d x} (22 cos (x))

\frac{d}{d x} (- 2 cos (x) sin (x))

y = (2 x - 1)^{5} (1 - x^{4})^{4}

\frac{2}{s \cdot ( s ^{2} + 4 s + 6 )}

What is the integral of (x^2)/((x+2)^2(x+4)^2) ?
The integral of (x^2)/((x+2)^2(x+4)^2) is -2ln|x+2|-1/(x+2)+2ln|x+4|-4/(x+4)+C