What is the integral of 1/(5sqrt(x)(6+\sqrt{x))} ?

The integral of 1/(5sqrt(x)(6+\sqrt{x))} is 2/5 ln|6+sqrt(x)|+C

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integral of 1/(5sqrt(x)(6+\sqrt{x))}

\int \frac{1}{5 x ( 6 + x )} d x

\frac{2}{5} ln 6 + x + C

Solution steps

\int \frac{1}{5 x ( 6 + x )} d x

Take the constant out:

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

= \frac{1}{5} \cdot \int \frac{1}{x ( 6 + x )} d x

Apply u-substitution

: \int \frac{2}{u} d u

= \frac{1}{5} \cdot \int \frac{2}{u} d u

Take the constant out:

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

= \frac{1}{5} \cdot 2 \cdot \int \frac{1}{u} d u

Use the common integral:

\int \frac{1}{u} d u = ln (∣ u ∣)

= \frac{1}{5} \cdot 2 ln ∣ u ∣

Substitute back

u = 6 + x

= \frac{1}{5} \cdot 2 ln 6 + x

Simplify

\frac{1}{5} \cdot 2 ln 6 + x : \frac{2}{5} ln 6 + x

= \frac{2}{5} ln 6 + x

Add a constant to the solution

= \frac{2}{5} ln 6 + x + C

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

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

- 4 cos (x)

12 y^{^{''}} - 12 y^{^{'}} + 6 y = 0

\frac{\partial}{\partial s} (\frac{( s - r )}{t})

What is the integral of 1/(5sqrt(x)(6+\sqrt{x))} ?
The integral of 1/(5sqrt(x)(6+\sqrt{x))} is 2/5 ln|6+sqrt(x)|+C