What is the integral of 1/(sqrt(9x^2+2)) ?

The integral of 1/(sqrt(9x^2+2)) is 1/3 ln| 3/(sqrt(2))x+sqrt(1/2 (2+9x^2))|+C

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integral of 1/(sqrt(9x^2+2))

\int \frac{1}{9 x ^{2} + 2} d x

\frac{1}{3} ln \frac{3}{2} x + \frac{1}{2} (2 + 9 x^{2}) + C

Solution steps

\int \frac{1}{9 x ^{2} + 2} d x

Apply Trigonometric Substitution

: \int \frac{sec ( u )}{3} d u

= \int \frac{sec ( u )}{3} d u

Take the constant out:

\int a \cdot f (x) d x = a \cdot \int f (x) d x

= \frac{1}{3} \cdot \int sec (u) d u

Use the common integral:

\int sec (u) d u = ln ∣ tan (u) + sec (u) ∣

= \frac{1}{3} ln ∣ tan (u) + sec (u) ∣

Substitute back

u = arctan (\frac{3}{2} x)

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

Simplify

\frac{1}{3} ln tan (arctan (\frac{3}{2} x)) + sec (arctan (\frac{3}{2} x)) : \frac{1}{3} ln \frac{3}{2} x + \frac{1}{2} (2 + 9 x^{2})

= \frac{1}{3} ln \frac{3}{2} x + \frac{1}{2} (2 + 9 x^{2})

Add a constant to the solution

= \frac{1}{3} ln \frac{3}{2} x + \frac{1}{2} (2 + 9 x^{2}) + C

\int (5 x + 6)^{- 1} d x

(3 x - 1) d x + (5 y + 7) d y = 0

(6, - 12), (15, - 3)

x \to 0.001 lim (\frac{e ^{x} - 1}{x})

x y \frac{d y}{d x} = ln (x)

What is the integral of 1/(sqrt(9x^2+2)) ?
The integral of 1/(sqrt(9x^2+2)) is 1/3 ln| 3/(sqrt(2))x+sqrt(1/2 (2+9x^2))|+C