What is the integral of 1/((x^2+4)(x+2)) ?

The integral of 1/((x^2+4)(x+2)) is 1/8 (-1/2 ln|x^2+4|+arctan(x/2))+1/8 ln|x+2|+C

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integral of 1/((x^2+4)(x+2))

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

\frac{1}{8} (- \frac{1}{2} ln x^{2} + 4 + arctan (\frac{x}{2})) + \frac{1}{8} ln ∣ x + 2 ∣ + C

Solution steps

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

Take the partial fraction of

\frac{1}{( x ^{2} + 4 ) ( x + 2 )} : \frac{- x + 2}{8 ( x ^{2} + 4 )} + \frac{1}{8 ( x + 2 )}

= \int \frac{- x + 2}{8 ( x ^{2} + 4 )} + \frac{1}{8 ( 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}{8 ( x ^{2} + 4 )} d x + \int \frac{1}{8 ( x + 2 )} d x

\int \frac{- x + 2}{8 ( x ^{2} + 4 )} d x = \frac{1}{8} (- \frac{1}{2} ln x^{2} + 4 + arctan (\frac{x}{2}))

\int \frac{1}{8 ( x + 2 )} d x = \frac{1}{8} ln ∣ x + 2 ∣

= \frac{1}{8} (- \frac{1}{2} ln x^{2} + 4 + arctan (\frac{x}{2})) + \frac{1}{8} ln ∣ x + 2 ∣

Add a constant to the solution

= \frac{1}{8} (- \frac{1}{2} ln x^{2} + 4 + arctan (\frac{x}{2})) + \frac{1}{8} ln ∣ x + 2 ∣ + C

t \to 0 lim (tan (7 t))

\frac{\partial}{\partial x} (10 x^{5} y^{4} e^{- x})

\frac{d}{d x} (e^{l o g_{3} (x)})

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\int_{- 6}^{6} (6 - ∣ x ∣) d x

What is the integral of 1/((x^2+4)(x+2)) ?
The integral of 1/((x^2+4)(x+2)) is 1/8 (-1/2 ln|x^2+4|+arctan(x/2))+1/8 ln|x+2|+C