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

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

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integral of x/(sqrt(64x^2-1))

\int \frac{x}{64 x ^{2} - 1} d x

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

Solution steps

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

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

Simplify

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

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

Add a constant to the solution

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

\frac{\partial}{\partial x} ((13 - x)^{80})

y (1 + x^{2}) y^{^{'}} = x (1 + y^{2})

\frac{\partial}{\partial x} (\frac{( y - b )}{m})

\int_{0}^{x} \frac{1}{x} d x

x \to - 1 + lim (\frac{x - 1}{x ^{3} - 1})

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