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

The integral of 2/(vsqrt(v^2-16)) is 1/2 arcsec(1/4 v)+C

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

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

\frac{1}{2} arcsec (\frac{1}{4} v) + C

Solution steps

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

Take the constant out:

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

= 2 \cdot \int \frac{1}{v v ^{2} - 16} d v

Apply Trigonometric Substitution

= 2 \cdot \int \frac{1}{4} d u

Take the constant out:

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

= 2 \cdot \frac{1}{4} \cdot \int 1 d u

Integral of a constant:

\int a d x = a x

= 2 \cdot \frac{1}{4} \cdot 1 \cdot u

Substitute back

u = arcsec (\frac{1}{4} v)

= 2 \cdot \frac{1}{4} \cdot 1 \cdot arcsec (\frac{1}{4} v)

Simplify

2 \cdot \frac{1}{4} \cdot 1 \cdot arcsec (\frac{1}{4} v) : \frac{1}{2} arcsec (\frac{1}{4} v)

= \frac{1}{2} arcsec (\frac{1}{4} v)

Add a constant to the solution

= \frac{1}{2} arcsec (\frac{1}{4} v) + C

What is the integral of 2/(vsqrt(v^2-16)) ?
The integral of 2/(vsqrt(v^2-16)) is 1/2 arcsec(1/4 v)+C