What is the integral of 3x^2sqrt(2x^3+7) ?

The integral of 3x^2sqrt(2x^3+7) is 1/3 (2x^3+7)^{3/2}+C

English

Español

Português

Français

Deutsch

Italiano

Русский

中文(简体)

한국어

日本語

Tiếng Việt

עברית

العربية

Popular >

integral of 3x^2sqrt(2x^3+7)

\int 3 x^{2} 2 x^{3} + 7 d x

\frac{1}{3} (2 x^{3} + 7)^{\frac{3}{2}} + C

Solution steps

\int 3 x^{2} 2 x^{3} + 7 d x

Take the constant out:

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

= 3 \cdot \int x^{2} 2 x^{3} + 7 d x

Apply u-substitution

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

= 3 \cdot \int \frac{u}{6} d u

Take the constant out:

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

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

Apply the Power Rule

: \frac{2}{3} u^{\frac{3}{2}}

= 3 \cdot \frac{1}{6} \cdot \frac{2}{3} u^{\frac{3}{2}}

Substitute back

u = 2 x^{3} + 7

= 3 \cdot \frac{1}{6} \cdot \frac{2}{3} (2 x^{3} + 7)^{\frac{3}{2}}

Simplify

3 \cdot \frac{1}{6} \cdot \frac{2}{3} (2 x^{3} + 7)^{\frac{3}{2}} : \frac{1}{3} (2 x^{3} + 7)^{\frac{3}{2}}

= \frac{1}{3} (2 x^{3} + 7)^{\frac{3}{2}}

Add a constant to the solution

= \frac{1}{3} (2 x^{3} + 7)^{\frac{3}{2}} + C

tan (x^{2})

\int e^{- s t} 2 t d t

\int 5 x \cdot e^{x} d x

\int (8 cos (x)) d x

\int \frac{x}{x ^{2} + 18 x + 48} d x

What is the integral of 3x^2sqrt(2x^3+7) ?
The integral of 3x^2sqrt(2x^3+7) is 1/3 (2x^3+7)^{3/2}+C