What is the integral of 1/(sqrt(2y^2+c)) ?

The integral of 1/(sqrt(2y^2+c)) is 1/(sqrt(2))ln|sqrt(2/c)y+sqrt((c+2y^2)/c)|+C

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integral of 1/(sqrt(2y^2+c))

\int \frac{1}{2 y ^{2} + c} d y

\frac{1}{2} ln \frac{2}{c} y + \frac{c + 2 y ^{2}}{c} + C

Solution steps

\int \frac{1}{2 y ^{2} + c} d y

Apply Trigonometric Substitution

: \int \frac{sec ( u )}{2} d u

= \int \frac{sec ( u )}{2} 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 sec (u) d u

Use the common integral:

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

= \frac{1}{2} ln ∣ tan (u) + sec (u) ∣

Substitute back

u = arctan (\frac{2}{c} y)

= \frac{1}{2} ln tan (arctan (\frac{2}{c} y)) + sec (arctan (\frac{2}{c} y))

Simplify

\frac{1}{2} ln tan (arctan (\frac{2}{c} y)) + sec (arctan (\frac{2}{c} y)) : \frac{1}{2} ln \frac{2}{c} y + \frac{c + 2 y ^{2}}{c}

= \frac{1}{2} ln \frac{2}{c} y + \frac{c + 2 y ^{2}}{c}

Add a constant to the solution

= \frac{1}{2} ln \frac{2}{c} y + \frac{c + 2 y ^{2}}{c} + C

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