What is the integral of 1/(sqrt(3x^2+12)) ?

The integral of 1/(sqrt(3x^2+12)) is 1/(sqrt(3))ln((|x+sqrt(4+x^2)|)/2)+C

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integral of 1/(sqrt(3x^2+12))

\int \frac{1}{3 x ^{2} + 12} d x

\frac{1}{3} ln (\frac{x + 4 + x ^{2}}{2}) + C

Solution steps

\int \frac{1}{3 x ^{2} + 12} 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 3} x)

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

Simplify

\frac{1}{3} ln tan (arctan (\frac{3}{2 3} x)) + sec (arctan (\frac{3}{2 3} x)) : \frac{1}{3} ln (\frac{x + 4 + x ^{2}}{2})

= \frac{1}{3} ln (\frac{x + 4 + x ^{2}}{2})

Add a constant to the solution

= \frac{1}{3} ln (\frac{x + 4 + x ^{2}}{2}) + C

\int \frac{2 x ^{3} + 1}{x ^{2} - 4 x} d x

\int \frac{1}{3 x ^{2} 25 - x ^{2}} d x

\int \frac{x}{5 - x} d x

\int (tan (θ) + sec (θ))^{2} d θ

\int \frac{x ^{2} - 2 x}{x} d x

What is the integral of 1/(sqrt(3x^2+12)) ?
The integral of 1/(sqrt(3x^2+12)) is 1/(sqrt(3))ln((|x+sqrt(4+x^2)|)/2)+C