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

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

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

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

- \frac{1}{2} 1 - 2 x^{2} + C

Solution steps

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

Apply u-substitution

: \int - \frac{1}{4 u} d u

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

Apply the Power Rule

: 2 u^{\frac{1}{2}}

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

Substitute back

u = 1 - 2 x^{2}

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

Simplify

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

= - \frac{1}{2} 1 - 2 x^{2}

Add a constant to the solution

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

\frac{d}{d x} (e^{xx} - e^{x})

f (x) = \frac{1}{x ^{6}} + \frac{1}{x ^{8}}

\int_{- 1}^{6} (x + 3)^{3} d x

sin (x), cos (x), [\frac{π}{4}, \frac{5 π}{4}]

x^{2} y^{^{''}} - 3 x y^{^{'}} + 13 y = 0

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