What is the integral of 3x\sqrt[3]{2x^2-11} ?

The integral of 3x\sqrt[3]{2x^2-11} is 9/16 (2x^2-11)^{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 3x\sqrt[3]{2x^2-11}

\int 3 x 3 2 x^{2} - 11 d x

\frac{9}{16} (2 x^{2} - 11)^{\frac{4}{3}} + C

Solution steps

\int 3 x 3 2 x^{2} - 11 d x

Take the constant out:

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

= 3 \cdot \int x 3 2 x^{2} - 11 d x

Apply u-substitution

: \int \frac{3 2 u - 11}{2} d u

= 3 \cdot \int \frac{3 2 u - 11}{2} d u

Take the constant out:

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

= 3 \cdot \frac{1}{2} \cdot \int 3 2 u - 11 d u

Apply u-substitution

: \int \frac{3 v}{2} d v

= 3 \cdot \frac{1}{2} \cdot \int \frac{3 v}{2} d v

Take the constant out:

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

= 3 \cdot \frac{1}{2} \cdot \frac{1}{2} \cdot \int 3 v d v

Apply the Power Rule

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

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

Substitute back

= 3 \cdot \frac{1}{2} \cdot \frac{1}{2} \cdot \frac{3}{4} (2 x^{2} - 11)^{\frac{4}{3}}

Simplify

3 \cdot \frac{1}{2} \cdot \frac{1}{2} \cdot \frac{3}{4} (2 x^{2} - 11)^{\frac{4}{3}} : \frac{9}{16} (2 x^{2} - 11)^{\frac{4}{3}}

= \frac{9}{16} (2 x^{2} - 11)^{\frac{4}{3}}

Add a constant to the solution

= \frac{9}{16} (2 x^{2} - 11)^{\frac{4}{3}} + C

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

\int (z + 2)^{3} d z

\int (2 x + 3) cos (x) d x

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

\int tan (θ) d θ

What is the integral of 3x\sqrt[3]{2x^2-11} ?
The integral of 3x\sqrt[3]{2x^2-11} is 9/16 (2x^2-11)^{4/3}+C