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

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

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

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

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

Solution steps

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

Apply Trigonometric Substitution

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

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

Take the constant out:

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

= \frac{1}{a} \cdot \int sec (u) d u

Use the common integral:

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

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

Substitute back

u = arctan (a y)

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

Simplify

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

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

Add a constant to the solution

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

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