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

The integral of x/(sqrt(3x-1)) is 2/9 (1/3 (3x-1)^{3/2}+sqrt(3x-1))+C

English

Español

Português

Français

Deutsch

Italiano

Русский

中文(简体)

한국어

日本語

Tiếng Việt

עברית

العربية

Popular >

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

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

\frac{2}{9} (\frac{1}{3} (3 x - 1)^{\frac{3}{2}} + 3 x - 1) + C

Solution steps

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

Apply u-substitution

: \int \frac{2 ( u ^{2} + 1 )}{9} d u

= \int \frac{2 ( u ^{2} + 1 )}{9} d u

Take the constant out:

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

= \frac{2}{9} \cdot \int u^{2} + 1 d u

Apply the Sum Rule:

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

= \frac{2}{9} (\int u^{2} d u + \int 1 d u)

\int u^{2} d u = \frac{u ^{3}}{3}

\int 1 d u = u

= \frac{2}{9} (\frac{u ^{3}}{3} + u)

Substitute back

u = 3 x - 1

= \frac{2}{9} (\frac{( 3 x - 1 ) ^{3}}{3} + 3 x - 1)

Simplify

\frac{2}{9} (\frac{( 3 x - 1 ) ^{3}}{3} + 3 x - 1) : \frac{2}{9} (\frac{1}{3} (3 x - 1)^{\frac{3}{2}} + 3 x - 1)

= \frac{2}{9} (\frac{1}{3} (3 x - 1)^{\frac{3}{2}} + 3 x - 1)

Add a constant to the solution

= \frac{2}{9} (\frac{1}{3} (3 x - 1)^{\frac{3}{2}} + 3 x - 1) + C

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

\int 5^{x^{4} + 2 x} (2 x^{3} + 1) d x

\int 3 x^{3} e^{x^{4}} d x

\int n a^{s i n (b x)} cos (b x) d x

\int \frac{x + 2}{x ^{2} + 2 x + 3} d x

What is the integral of x/(sqrt(3x-1)) ?
The integral of x/(sqrt(3x-1)) is 2/9 (1/3 (3x-1)^{3/2}+sqrt(3x-1))+C