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

The integral of 1/(sqrt(x)(x+2)) is sqrt(2)arctan(sqrt(x/2))+C

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

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

2 arctan (\frac{x}{2}) + C

Solution steps

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

Apply u-substitution

= \int \frac{2}{u ^{2} + 2} d u

Take the constant out:

\int a \cdot f (x) d x = a \cdot \int f (x) d x

= 2 \cdot \int \frac{1}{u ^{2} + 2} d u

Apply Integral Substitution

= 2 \cdot \int \frac{1}{2 ( v ^{2} + 1 )} d v

Take the constant out:

\int a \cdot f (x) d x = a \cdot \int f (x) d x

= 2 \cdot \frac{1}{2} \cdot \int \frac{1}{v ^{2} + 1} d v

Use the common integral:

\int \frac{1}{v ^{2} + 1} d v = arctan (v)

= 2 \cdot \frac{1}{2} arctan (v)

Substitute back

= 2 \cdot \frac{1}{2} arctan (\frac{x}{2})

Simplify

2 \cdot \frac{1}{2} arctan (\frac{x}{2}) : 2 arctan (\frac{x}{2})

= 2 arctan (\frac{x}{2})

Add a constant to the solution

= 2 arctan (\frac{x}{2}) + C

What is the integral of 1/(sqrt(x)(x+2)) ?
The integral of 1/(sqrt(x)(x+2)) is sqrt(2)arctan(sqrt(x/2))+C