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

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

English

Español

Português

Français

Deutsch

Italiano

Русский

中文(简体)

한국어

日本語

Tiếng Việt

עברית

العربية

Popular >

integral of (x^3)/(sqrt(3-x^2))

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

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

Solution steps

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

Apply u-substitution

: \int u^{2} - 3 d u

= \int u^{2} - 3 d u

Apply the Sum Rule:

\int f (x) \pm g (x) d x = \int f (x) d x \pm \int g (x) d x

= \int u^{2} d u - \int 3 d u

\int u^{2} d u = \frac{u ^{3}}{3}

\int 3 d u = 3 u

= \frac{u ^{3}}{3} - 3 u

Substitute back

u = 3 - x^{2}

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

Simplify

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

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

Add a constant to the solution

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

\int (x - 1)^{- 1} d x

\int \frac{- 2 x + 6}{( x + 1 ) ( x ^{2} + x + 1 )} d x

\int \frac{x ^{2}}{( x ^{2} + 9 ) ^{\frac{3}{2}}} d x

\int (\frac{y ^{4} - 1}{y})^{2} d y

\int cos^{8} (x) sin^{3} (x) d x

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