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

y^{^{'}} + tan (x) y - sin (x) = 0

\int \frac{1}{x} + 2 x d x

\frac{\partial}{\partial x} (4 sin (4 x))

\frac{d}{d x} (arccos (x + y))

f (x) = 3 x^{2}, at x = (6)

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