What is the integral of (6x^2)/(5sqrt(4x^3-3)) ?

The integral of (6x^2)/(5sqrt(4x^3-3)) is 1/5 sqrt(4x^3-3)+C

English

Español

Português

Français

Deutsch

Italiano

Русский

中文(简体)

한국어

日本語

Tiếng Việt

עברית

العربية

Popular >

integral of (6x^2)/(5sqrt(4x^3-3))

\int \frac{6 x ^{2}}{5 4 x ^{3} - 3} d x

\frac{1}{5} 4 x^{3} - 3 + C

Solution steps

\int \frac{6 x ^{2}}{5 4 x ^{3} - 3} d x

Take the constant out:

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

= \frac{6}{5} \cdot \int \frac{x ^{2}}{4 x ^{3} - 3} d x

Apply u-substitution

: \int \frac{u}{2 x 4 u ^{\frac{3}{2}} - 3} d u

= \frac{6}{5} \cdot \int \frac{1}{3 4 u - 3} d u

Take the constant out:

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

= \frac{6}{5} \cdot \frac{1}{3} \cdot \int \frac{1}{4 u - 3} d u

Apply u-substitution

: \int \frac{1}{4 v} d v

= \frac{6}{5} \cdot \frac{1}{3} \cdot \int \frac{1}{4 v} d v

Take the constant out:

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

= \frac{6}{5} \cdot \frac{1}{3} \cdot \frac{1}{4} \cdot \int \frac{1}{v} d v

Apply the Power Rule

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

= \frac{6}{5} \cdot \frac{1}{3} \cdot \frac{1}{4} \cdot 2 v^{\frac{1}{2}}

Substitute back

= \frac{6}{5} \cdot \frac{1}{3} \cdot \frac{1}{4} \cdot 2 (4 x^{3} - 3)^{\frac{1}{2}}

Simplify

\frac{6}{5} \cdot \frac{1}{3} \cdot \frac{1}{4} \cdot 2 (4 x^{3} - 3)^{\frac{1}{2}} : \frac{1}{5} 4 x^{3} - 3

= \frac{1}{5} 4 x^{3} - 3

Add a constant to the solution

= \frac{1}{5} 4 x^{3} - 3 + C

\int \frac{x - 5}{x ^{2} - x - 6} d x

\int \frac{3 x ^{2} - 5 x + 7}{x + 1} d x

\int (\frac{1}{3 x} - 3 x^{\frac{5}{2}}) d x

\int \frac{6 x ^{5} - x + 5 x ^{2}}{x ^{3}} d x

\int e^{- p x} sin (w x) d x

What is the integral of (6x^2)/(5sqrt(4x^3-3)) ?
The integral of (6x^2)/(5sqrt(4x^3-3)) is 1/5 sqrt(4x^3-3)+C