What is the integral of 1/(6sqrt(x)(3+\sqrt{x))} ?

The integral of 1/(6sqrt(x)(3+\sqrt{x))} is 1/3 ln|3+sqrt(x)|+C

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integral of 1/(6sqrt(x)(3+\sqrt{x))}

\int \frac{1}{6 x ( 3 + x )} d x

\frac{1}{3} ln 3 + x + C

Solution steps

\int \frac{1}{6 x ( 3 + x )} d x

Take the constant out:

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

= \frac{1}{6} \cdot \int \frac{1}{x ( 3 + x )} d x

Apply u-substitution

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

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

Use the common integral:

\int \frac{1}{u} d u = ln (∣ u ∣)

= \frac{1}{6} \cdot 2 ln ∣ u ∣

Substitute back

u = 3 + x

= \frac{1}{6} \cdot 2 ln 3 + x

Simplify

\frac{1}{6} \cdot 2 ln 3 + x : \frac{1}{3} ln 3 + x

= \frac{1}{3} ln 3 + x

Add a constant to the solution

= \frac{1}{3} ln 3 + x + C

(x^{2} - x y + 2 y^{2}) d x + (x^{2} - 2 x y) d y = 0

lo g_{2} (3 - 2 x)

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

x \to 0 lim (1 + csc (x))

\int (sec (x) + tan (x)) d x

What is the integral of 1/(6sqrt(x)(3+\sqrt{x))} ?
The integral of 1/(6sqrt(x)(3+\sqrt{x))} is 1/3 ln|3+sqrt(x)|+C