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

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

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

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

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

Solution steps

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

Take the partial fraction of

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

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

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

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

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

Add a constant to the solution

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

θ^{^{''}} + \frac{a}{b} \cdot θ = 0

f (x) = x e^{4 x + s i n (x) - 10}

\frac{\partial}{\partial x} (2 e^{x yz} sin (x w))

\int_{2}^{5} \frac{1}{x} d x

x^{2} - 1

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