What is the integral of (x^3)/(sqrt(1-2x^2)) ?

The integral of (x^3)/(sqrt(1-2x^2)) is -1/6 (x^2+1)sqrt(-2x^2+1)+C

English

Español

Português

Français

Deutsch

Italiano

Русский

中文(简体)

한국어

日本語

Tiếng Việt

עברית

العربية

integral of (x^3)/(sqrt(1-2x^2))

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

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

Solution steps

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

Apply u-substitution

= \int \frac{u ^{2} - 1}{4} d u

Take the constant out:

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

= \frac{1}{4} \cdot \int u^{2} - 1 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

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

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

\int 1 d u = u

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

Substitute back

u = 1 - 2 x^{2}

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

Simplify

\frac{1}{4} (\frac{( 1 - 2 x ^{2} ) ^{3}}{3} - 1 - 2 x^{2}) : - \frac{1}{6} (x^{2} + 1) - 2 x^{2} + 1

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

Add a constant to the solution

= - \frac{1}{6} (x^{2} + 1) - 2 x^{2} + 1 + C

What is the integral of (x^3)/(sqrt(1-2x^2)) ?
The integral of (x^3)/(sqrt(1-2x^2)) is -1/6 (x^2+1)sqrt(-2x^2+1)+C