What is the integral of 4/(sqrt(x)(x+6)) ?

The integral of 4/(sqrt(x)(x+6)) is (4sqrt(2))/(sqrt(3))arctan(sqrt(x/6))+C

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integral of 4/(sqrt(x)(x+6))

\int \frac{4}{x ( x + 6 )} d x

\frac{4 2}{3} arctan (\frac{x}{6}) + C

Solution steps

\int \frac{4}{x ( x + 6 )} d x

Take the constant out:

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

= 4 \cdot \int \frac{1}{x ( x + 6 )} d x

Apply u-substitution

: \int \frac{2}{u ^{2} + 6} d u

= 4 \cdot \int \frac{2}{u ^{2} + 6} d u

Take the constant out:

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

= 4 \cdot 2 \cdot \int \frac{1}{u ^{2} + 6} d u

Apply Integral Substitution

= 4 \cdot 2 \cdot \int \frac{1}{6 ( v ^{2} + 1 )} d v

Take the constant out:

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

= 4 \cdot 2 \cdot \frac{1}{6} \cdot \int \frac{1}{v ^{2} + 1} d v

Use the common integral:

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

= 4 \cdot 2 \cdot \frac{1}{6} arctan (v)

Substitute back

= 4 \cdot 2 \cdot \frac{1}{6} arctan (\frac{x}{6})

Simplify

4 \cdot 2 \cdot \frac{1}{6} arctan (\frac{x}{6}) : \frac{4 2}{3} arctan (\frac{x}{6})

= \frac{4 2}{3} arctan (\frac{x}{6})

Add a constant to the solution

= \frac{4 2}{3} arctan (\frac{x}{6}) + C

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\frac{\partial}{\partial x} (4 x^{2} y^{2})

(- 2, 1), (3, - 9)

\int \frac{1}{( r ^{2} + x ^{2} ) ^{\frac{3}{2}}} d x

What is the integral of 4/(sqrt(x)(x+6)) ?
The integral of 4/(sqrt(x)(x+6)) is (4sqrt(2))/(sqrt(3))arctan(sqrt(x/6))+C