What is the integral of x/(\sqrt[3]{1-2x^2)} ?

The integral of x/(\sqrt[3]{1-2x^2)} is -3/8 (1-2x^2)^{2/3}+C

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integral of x/(\sqrt[3]{1-2x^2)}

- \frac{3}{8} (1 - 2 x^{2})^{\frac{2}{3}} + C

Solution steps

Apply u-substitution

Take the constant out:

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

Apply the Power Rule

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

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

Substitute back

u = 1 - 2 x^{2}

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

Simplify

- \frac{1}{4} \cdot \frac{3}{2} (1 - 2 x^{2})^{\frac{2}{3}} : - \frac{3}{8} (1 - 2 x^{2})^{\frac{2}{3}}

= - \frac{3}{8} (1 - 2 x^{2})^{\frac{2}{3}}

Add a constant to the solution

= - \frac{3}{8} (1 - 2 x^{2})^{\frac{2}{3}} + C

\int t ln (t + 4) d t

\frac{d}{d x} (\frac{36 x}{x ^{2} + 36})

f (x) = (4 + 5 x)^{2}, (- 3, 121)

f (x) = 2 x, (18, 6)

\frac{d y}{d x} = \frac{e ^{y} + 1}{e ^{y}}

What is the integral of x/(\sqrt[3]{1-2x^2)} ?
The integral of x/(\sqrt[3]{1-2x^2)} is -3/8 (1-2x^2)^{2/3}+C