What is the integral of x/(sqrt(9-64x^2)) ?

The integral of x/(sqrt(9-64x^2)) is -1/64 sqrt(9-64x^2)+C

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integral of x/(sqrt(9-64x^2))

\int \frac{x}{9 - 64 x ^{2}} d x

- \frac{1}{64} 9 - 64 x^{2} + C

Solution steps

\int \frac{x}{9 - 64 x ^{2}} d x

Apply u-substitution

: \int - \frac{1}{128 u} d u

= \int - \frac{1}{128 u} d u

Take the constant out:

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

= - \frac{1}{128} \cdot \int \frac{1}{u} d u

Apply the Power Rule

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

= - \frac{1}{128} \cdot 2 u^{\frac{1}{2}}

Substitute back

u = 9 - 64 x^{2}

= - \frac{1}{128} \cdot 2 (9 - 64 x^{2})^{\frac{1}{2}}

Simplify

- \frac{1}{128} \cdot 2 (9 - 64 x^{2})^{\frac{1}{2}} : - \frac{1}{64} 9 - 64 x^{2}

= - \frac{1}{64} 9 - 64 x^{2}

Add a constant to the solution

= - \frac{1}{64} 9 - 64 x^{2} + C

\int_{0}^{16} 1 + 3 x d x

\frac{d}{d x} (ln (x + y - 1))

x^{2} (\frac{d y}{d x}) = y - y x

y = x^{2} - 6 x, (5, - 5)

y^{^{''}} - 2 y^{^{'}} + 10 y = 9 e t csc (3 t)

What is the integral of x/(sqrt(9-64x^2)) ?
The integral of x/(sqrt(9-64x^2)) is -1/64 sqrt(9-64x^2)+C