What is the integral of 1/(sqrt(x^2+14)) ?

The integral of 1/(sqrt(x^2+14)) is ln| 1/(sqrt(14))x+sqrt(1/14 (14+x^2))|+C

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integral of 1/(sqrt(x^2+14))

\int \frac{1}{x ^{2} + 14} d x

ln \frac{1}{14} x + \frac{1}{14} (14 + x^{2}) + C

Solution steps

\int \frac{1}{x ^{2} + 14} 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}{14} x)

= ln tan (arctan (\frac{1}{14} x)) + sec (arctan (\frac{1}{14} x))

Simplify

ln tan (arctan (\frac{1}{14} x)) + sec (arctan (\frac{1}{14} x)) : ln \frac{1}{14} x + \frac{1}{14} (14 + x^{2})

= ln \frac{1}{14} x + \frac{1}{14} (14 + x^{2})

Add a constant to the solution

= ln \frac{1}{14} x + \frac{1}{14} (14 + x^{2}) + C

What is the integral of 1/(sqrt(x^2+14)) ?
The integral of 1/(sqrt(x^2+14)) is ln| 1/(sqrt(14))x+sqrt(1/14 (14+x^2))|+C