What is the integral of 1/(sqrt(3x)(1+3x)) ?

The integral of 1/(sqrt(3x)(1+3x)) is 2/3 arctan(sqrt(3)sqrt(x))+C

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integral of 1/(sqrt(3x)(1+3x))

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

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

Solution steps

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

3 x = 3 x

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

Take the constant out:

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

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

Apply u-substitution

: \int \frac{2}{1 + 3 u ^{2}} d u

= \frac{1}{3} \cdot \int \frac{2}{1 + 3 u ^{2}} d u

Take the constant out:

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

= \frac{1}{3} \cdot 2 \cdot \int \frac{1}{1 + 3 u ^{2}} d u

Apply Integral Substitution

= \frac{1}{3} \cdot 2 \cdot \int \frac{1}{3 ( 1 + v ^{2} )} d v

Take the constant out:

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

= \frac{1}{3} \cdot 2 \cdot \frac{1}{3} \cdot \int \frac{1}{1 + v ^{2}} d v

Use the common integral:

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

= \frac{1}{3} \cdot 2 \cdot \frac{1}{3} arctan (v)

Substitute back

= \frac{1}{3} \cdot 2 \cdot \frac{1}{3} arctan (3 x)

Simplify

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

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

Add a constant to the solution

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

\int \frac{x + 2}{x - 2} (x^{2} - 4 x) d x

n = 2 \sum \infty \frac{1}{2 n ln ( n )}

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

\frac{d}{d x} (x (a cos (x) + b sin (x)))

\int_{- 1}^{e - 2} \frac{t}{t ^{2} + 2} d t

What is the integral of 1/(sqrt(3x)(1+3x)) ?
The integral of 1/(sqrt(3x)(1+3x)) is 2/3 arctan(sqrt(3)sqrt(x))+C