What is the integral of x/(sqrt(1-16x^2)) ?

The integral of x/(sqrt(1-16x^2)) is -1/16 sqrt(1-16x^2)+C

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integral of x/(sqrt(1-16x^2))

\int \frac{x}{1 - 16 x ^{2}} d x

- \frac{1}{16} 1 - 16 x^{2} + C

Solution steps

\int \frac{x}{1 - 16 x ^{2}} d x

Apply u-substitution

: \int - \frac{1}{32 u} d u

= \int - \frac{1}{32 u} d u

Take the constant out:

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

= - \frac{1}{32} \cdot \int \frac{1}{u} d u

Apply the Power Rule

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

= - \frac{1}{32} \cdot 2 u^{\frac{1}{2}}

Substitute back

u = 1 - 16 x^{2}

= - \frac{1}{32} \cdot 2 (1 - 16 x^{2})^{\frac{1}{2}}

Simplify

- \frac{1}{32} \cdot 2 (1 - 16 x^{2})^{\frac{1}{2}} : - \frac{1}{16} 1 - 16 x^{2}

= - \frac{1}{16} 1 - 16 x^{2}

Add a constant to the solution

= - \frac{1}{16} 1 - 16 x^{2} + C

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What is the integral of x/(sqrt(1-16x^2)) ?
The integral of x/(sqrt(1-16x^2)) is -1/16 sqrt(1-16x^2)+C