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

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

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

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

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

Solution steps

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

Apply u-substitution

: \int - \frac{1}{6 u} d u

= \int - \frac{1}{6 u} d u

Take the constant out:

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

= - \frac{1}{6} \cdot \int \frac{1}{u} d u

Apply the Power Rule

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

= - \frac{1}{6} \cdot 2 u^{\frac{1}{2}}

Substitute back

u = 1 - 3 x^{2}

= - \frac{1}{6} \cdot 2 (1 - 3 x^{2})^{\frac{1}{2}}

Simplify

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

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

Add a constant to the solution

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

f (x) = x^{2}, g (x) = 9 + x

f (x) = 1 8^{x}

\int \frac{y}{y ^{2} + 2} d y

\int (10 x - 3) d x

x^{2} - 3 x + 2, - 3 x + 3

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