What is the integral of (x/(sqrt(1-4x^2))) ?

The integral of (x/(sqrt(1-4x^2))) is -1/4 sqrt(1-4x^2)+C

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integral of (x/(sqrt(1-4x^2)))

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

- \frac{1}{4} 1 - 4 x^{2} + C

Solution steps

\int \frac{x}{1 - 4 x ^{2}} d x

Apply u-substitution

: \int - \frac{1}{8 u} d u

= \int - \frac{1}{8 u} d u

Take the constant out:

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

= - \frac{1}{8} \cdot \int \frac{1}{u} d u

Apply the Power Rule

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

= - \frac{1}{8} \cdot 2 u^{\frac{1}{2}}

Substitute back

u = 1 - 4 x^{2}

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

Simplify

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

= - \frac{1}{4} 1 - 4 x^{2}

Add a constant to the solution

= - \frac{1}{4} 1 - 4 x^{2} + C

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What is the integral of (x/(sqrt(1-4x^2))) ?
The integral of (x/(sqrt(1-4x^2))) is -1/4 sqrt(1-4x^2)+C