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

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

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

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

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

Solution steps

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

Apply u-substitution

= \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

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

Substitute back

u = 16 x^{2} - 1

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

Simplify

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

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

Add a constant to the solution

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

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