What is the integral of 1/(sqrt(x^2+6)) ?

The integral of 1/(sqrt(x^2+6)) is ln| 1/(sqrt(6))x+sqrt(1/6 (6+x^2))|+C

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integral of 1/(sqrt(x^2+6))

\int \frac{1}{x ^{2} + 6} d x

ln \frac{1}{6} x + \frac{1}{6} (6 + x^{2}) + C

Solution steps

\int \frac{1}{x ^{2} + 6} 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}{6} x)

= ln tan (arctan (\frac{1}{6} x)) + sec (arctan (\frac{1}{6} x))

Simplify

ln tan (arctan (\frac{1}{6} x)) + sec (arctan (\frac{1}{6} x)) : ln \frac{1}{6} x + \frac{1}{6} (6 + x^{2})

= ln \frac{1}{6} x + \frac{1}{6} (6 + x^{2})

Add a constant to the solution

= ln \frac{1}{6} x + \frac{1}{6} (6 + x^{2}) + C

\frac{\partial}{\partial x} (\frac{2 x - 3 t}{3 x + 5 t})

\int \frac{6}{x ( ln ( x ) ) ^{2}} d x

3 t^{2} - 3

\int \frac{y ^{2} - 25}{y ^{3}} d y

\int \frac{1}{600 - x} d x

What is the integral of 1/(sqrt(x^2+6)) ?
The integral of 1/(sqrt(x^2+6)) is ln| 1/(sqrt(6))x+sqrt(1/6 (6+x^2))|+C