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

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

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

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

ln \frac{x}{a} + \frac{a + x ^{2}}{a} + C

Solution steps

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

Apply Trigonometric Substitution

= \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} x)

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

Simplify

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

= ln \frac{x}{a} + \frac{a + x ^{2}}{a}

Add a constant to the solution

= ln \frac{x}{a} + \frac{a + x ^{2}}{a} + C

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