What is the integral of (2x^3-1)/(sqrt(3x^4-6x+9)) ?

The integral of (2x^3-1)/(sqrt(3x^4-6x+9)) is 1/3 sqrt(3x^4-6x+9)+C

English

Español

Português

Français

Deutsch

Italiano

Русский

中文(简体)

한국어

日本語

Tiếng Việt

עברית

العربية

Popular >

integral of (2x^3-1)/(sqrt(3x^4-6x+9))

\int \frac{2 x ^{3} - 1}{3 x ^{4} - 6 x + 9} d x

\frac{1}{3} 3 x^{4} - 6 x + 9 + C

Solution steps

\int \frac{2 x ^{3} - 1}{3 x ^{4} - 6 x + 9} d x

Apply u-substitution

: \int \frac{1}{6 u} d u

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

Apply the Power Rule

: 2 u^{\frac{1}{2}}

= \frac{1}{6} \cdot 2 u^{\frac{1}{2}}

Substitute back

u = 3 x^{4} - 6 x + 9

= \frac{1}{6} \cdot 2 (3 x^{4} - 6 x + 9)^{\frac{1}{2}}

Simplify

\frac{1}{6} \cdot 2 (3 x^{4} - 6 x + 9)^{\frac{1}{2}} : \frac{1}{3} 3 x^{4} - 6 x + 9

= \frac{1}{3} 3 x^{4} - 6 x + 9

Add a constant to the solution

= \frac{1}{3} 3 x^{4} - 6 x + 9 + C

\int cos (\frac{x}{3}) sin (\frac{x}{2}) d x

\int \frac{a}{1 - cos ( b x )} d x

\int 200 - 0.02 x d x

\int x \cdot sec^{2} (4 x) d x

What is the integral of (2x^3-1)/(sqrt(3x^4-6x+9)) ?
The integral of (2x^3-1)/(sqrt(3x^4-6x+9)) is 1/3 sqrt(3x^4-6x+9)+C