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

The integral of 3/(1+sqrt(3x)) is 2sqrt(3)sqrt(x)-2ln|sqrt(3)sqrt(x)+1|+3+C

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

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

23 x - 2 ln 3 x + 1 + 3 + C

Solution steps

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

Take the constant out:

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

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

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

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

Apply the Sum Rule:

\int f (x) \pm g (x) d x = \int f (x) d x \pm \int g (x) d x

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

\int \frac{3 x}{1 + 3 x} d x = \frac{1}{3} (3 x - 23 x + 2 ln 3 x + 1 - 3)

\int 1 d x = x

= 3 (- \frac{1}{3} (3 x - 23 x + 2 ln 3 x + 1 - 3) + x)

Simplify

3 (- \frac{1}{3} (3 x - 23 x + 2 ln 3 x + 1 - 3) + x) : 23 x - 2 ln 3 x + 1 + 3

= 23 x - 2 ln 3 x + 1 + 3

Add a constant to the solution

= 23 x - 2 ln 3 x + 1 + 3 + C

\frac{\partial}{\partial x} (6 x^{2} y)

d y = \frac{x + 6}{x + 1} d x

y = 4 sin (5 x - 2)

(t^{2} + 1) \frac{d y}{d t} = y t - y

n = 1 \sum \infty \frac{1}{n 8 ^{n}}

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