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

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

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integral of v/(sqrt(v^2-16))

\int \frac{v}{v ^{2} - 16} d v

v^{2} - 16 + C

Solution steps

\int \frac{v}{v ^{2} - 16} d v

Apply u-substitution

: \int \frac{1}{2 u} d u

= \int \frac{1}{2 u} d u

Take the constant out:

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

= \frac{1}{2} \cdot \int \frac{1}{u} d u

Apply the Power Rule

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

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

Substitute back

u = v^{2} - 16

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

Simplify

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

= v^{2} - 16

Add a constant to the solution

= v^{2} - 16 + C

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