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

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

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

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

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

Solution steps

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

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

Simplify

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

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

Add a constant to the solution

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

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