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

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

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

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

\frac{1}{2} (\frac{1}{2} ln \frac{x ^{2}}{4} + 1 - arctan (\frac{x}{2})) + \frac{1}{2} ln ∣ x + 2 ∣ + C

Solution steps

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

Take the partial fraction of

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

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

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

\int \frac{1}{2 ( x + 2 )} d x = \frac{1}{2} ln ∣ x + 2 ∣

= \frac{1}{2} (\frac{1}{2} ln \frac{x ^{2}}{4} + 1 - arctan (\frac{x}{2})) + \frac{1}{2} ln ∣ x + 2 ∣

Add a constant to the solution

= \frac{1}{2} (\frac{1}{2} ln \frac{x ^{2}}{4} + 1 - arctan (\frac{x}{2})) + \frac{1}{2} ln ∣ x + 2 ∣ + C

b + x^{2} - k ln (\frac{( x + 1 ) ( x - 1 )}{k}) - 1

\frac{2 x}{1 + x ^{2}}

\int \frac{x + 9}{2 x ^{3} - 18 x} d x

2 \frac{d y}{d x} + 18 y = 9

\int 5 sin^{2} (3 x) d x

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