What is the integral of (2x-3)\sqrt[3]{(2x^2-6x+7)} ?

The integral of (2x-3)\sqrt[3]{(2x^2-6x+7)} is 3/(16\sqrt[3]{2)}(4x^2-12x+14)^{4/3}+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)\sqrt[3]{(2x^2-6x+7)}

Solution steps

Apply u-substitution

Take the constant out:

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

Apply u-substitution

Take the constant out:

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

Apply the Power Rule

: \frac{3}{4} v^{\frac{4}{3}}

Substitute back

Simplify

Add a constant to the solution

\int \frac{15}{x x ^{2} - 1} d x

sin (x) cos^{2} (x)

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

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

\int (2 t + 2) d t

What is the integral of (2x-3)\sqrt[3]{(2x^2-6x+7)} ?
The integral of (2x-3)\sqrt[3]{(2x^2-6x+7)} is 3/(16\sqrt[3]{2)}(4x^2-12x+14)^{4/3}+C