What is the integral of 1/(sqrt(a^2+v^2)) ?

The integral of 1/(sqrt(a^2+v^2)) is ln((|v+sqrt(a^2+v^2)|)/(|a|))+C

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integral of 1/(sqrt(a^2+v^2))

\int \frac{1}{a ^{2} + v ^{2}} d v

ln (\frac{v + a ^{2} + v ^{2}}{∣ a ∣}) + C

Solution steps

\int \frac{1}{a ^{2} + v ^{2}} d v

Apply Trigonometric Substitution

: \int sec (u) d u

= \int sec (u) d u

Use the common integral:

\int sec (u) d u = ln ∣ tan (u) + sec (u) ∣

= ln ∣ tan (u) + sec (u) ∣

Substitute back

u = arctan (\frac{1}{a} v)

= ln tan (arctan (\frac{1}{a} v)) + sec (arctan (\frac{1}{a} v))

Simplify

ln tan (arctan (\frac{1}{a} v)) + sec (arctan (\frac{1}{a} v)) : ln (\frac{v + a ^{2} + v ^{2}}{∣ a ∣})

= ln (\frac{v + a ^{2} + v ^{2}}{∣ a ∣})

Add a constant to the solution

= ln (\frac{v + a ^{2} + v ^{2}}{∣ a ∣}) + C

\frac{d}{d x} (\frac{3}{4} x^{\frac{4}{3}})

n = 1 \sum \infty \frac{6 n}{2 n + 1}

e^{1 - s}

ln (arctan (x + 1))

\frac{\partial}{\partial t} (\frac{rs}{s ^{2} + t ^{2}})

What is the integral of 1/(sqrt(a^2+v^2)) ?
The integral of 1/(sqrt(a^2+v^2)) is ln((|v+sqrt(a^2+v^2)|)/(|a|))+C