What is the integral of 1/(4sqrt(x)(6+\sqrt{x))} ?

The integral of 1/(4sqrt(x)(6+\sqrt{x))} is 1/2 ln|6+sqrt(x)|+C

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integral of 1/(4sqrt(x)(6+\sqrt{x))}

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

\frac{1}{2} ln 6 + x + C

Solution steps

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

Apply u-substitution

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

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

Use the common integral:

\int \frac{1}{u} d u = ln (∣ u ∣)

= \frac{1}{4} \cdot 2 ln ∣ u ∣

Substitute back

u = 6 + x

= \frac{1}{4} \cdot 2 ln 6 + x

Simplify

\frac{1}{4} \cdot 2 ln 6 + x : \frac{1}{2} ln 6 + x

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

Add a constant to the solution

= \frac{1}{2} ln 6 + x + C

\frac{2000 t}{4 t + 75}

\frac{\partial}{\partial x} (x + 2)

\int \frac{15 e ^{5 z}}{e ^{5 z} + 2} d z

\frac{\partial}{\partial t} (x^{2} y cos (\frac{z}{t}))

\int sec^{5} (x) tan (x) d x

What is the integral of 1/(4sqrt(x)(6+\sqrt{x))} ?
The integral of 1/(4sqrt(x)(6+\sqrt{x))} is 1/2 ln|6+sqrt(x)|+C