What is the integral of 1/(sqrt(x^2+3)) ?

The integral of 1/(sqrt(x^2+3)) is ln| 1/(sqrt(3))x+sqrt(1/3 (3+x^2))|+C

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integral of 1/(sqrt(x^2+3))

\int \frac{1}{x ^{2} + 3} d x

ln \frac{1}{3} x + \frac{1}{3} (3 + x^{2}) + C

Solution steps

\int \frac{1}{x ^{2} + 3} d x

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}{3} x)

= ln tan (arctan (\frac{1}{3} x)) + sec (arctan (\frac{1}{3} x))

Simplify

ln tan (arctan (\frac{1}{3} x)) + sec (arctan (\frac{1}{3} x)) : ln \frac{1}{3} x + \frac{1}{3} (3 + x^{2})

= ln \frac{1}{3} x + \frac{1}{3} (3 + x^{2})

Add a constant to the solution

= ln \frac{1}{3} x + \frac{1}{3} (3 + x^{2}) + C

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What is the integral of 1/(sqrt(x^2+3)) ?
The integral of 1/(sqrt(x^2+3)) is ln| 1/(sqrt(3))x+sqrt(1/3 (3+x^2))|+C