What is the integral of 1/(sqrt(3x^2+7)) ?

The integral of 1/(sqrt(3x^2+7)) is 1/(sqrt(3))ln|sqrt(3/7)x+sqrt(1/7 (7+3x^2))|+C

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integral of 1/(sqrt(3x^2+7))

\int \frac{1}{3 x ^{2} + 7} d x

\frac{1}{3} ln \frac{3}{7} x + \frac{1}{7} (7 + 3 x^{2}) + C

Solution steps

\int \frac{1}{3 x ^{2} + 7} d x

Apply Trigonometric Substitution

= \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}{7} x)

= \frac{1}{3} ln tan (arctan (\frac{3}{7} x)) + sec (arctan (\frac{3}{7} x))

Simplify

\frac{1}{3} ln tan (arctan (\frac{3}{7} x)) + sec (arctan (\frac{3}{7} x)) : \frac{1}{3} ln \frac{3}{7} x + \frac{1}{7} (7 + 3 x^{2})

= \frac{1}{3} ln \frac{3}{7} x + \frac{1}{7} (7 + 3 x^{2})

Add a constant to the solution

= \frac{1}{3} ln \frac{3}{7} x + \frac{1}{7} (7 + 3 x^{2}) + C

What is the integral of 1/(sqrt(3x^2+7)) ?
The integral of 1/(sqrt(3x^2+7)) is 1/(sqrt(3))ln|sqrt(3/7)x+sqrt(1/7 (7+3x^2))|+C