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

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

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

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

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

Solution steps

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

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} y)

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

Simplify

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

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

Add a constant to the solution

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

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

x \to - 3 lim (- x + 6 - 3)

(3 x^{2} + 1)^{2}

\frac{d}{d x} (- 4 sin (x) - 9 ln (x) + 11)

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