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

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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

= \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

3 x e^{3 x}

\frac{d}{d y} (y e^{x y})

6 \frac{d y}{d t} - 2 y = t y^{4}, y (0) = - 2

\frac{d}{d t} (7 sin (t))

\frac{t}{( t - 3 ) ^{2}}

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