What is the integral of 2xsqrt(3-2x^2) ?

The integral of 2xsqrt(3-2x^2) is -1/3 (3-2x^2)^{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 2xsqrt(3-2x^2)

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

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

Solution steps

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

Take the constant out:

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

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

Apply u-substitution

: \int - \frac{u}{4} d u

= 2 \cdot \int - \frac{u}{4} d u

Take the constant out:

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

= 2 (- \frac{1}{4} \cdot \int u d u)

Apply the Power Rule

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

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

Substitute back

u = 3 - 2 x^{2}

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

Simplify

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

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

Add a constant to the solution

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

\int - 5 x + 10 - e^{- x} d x

\int (1 + sin^{2} (θ) csc (θ)) d θ

\int \frac{x ^{3}}{x ^{2} + 22 x + 121} d x

\int \frac{2}{( t + 3 ) t ^{2} + 6 t - 7} d t

\int 7 x^{8} - 0^{2} 2 x^{3} d x

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