What is the integral of 1/(sqrt(36x^2+2)) ?

The integral of 1/(sqrt(36x^2+2)) is 1/6 ln|3sqrt(2)x+sqrt(1+18x^2)|+C

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integral of 1/(sqrt(36x^2+2))

\int \frac{1}{36 x ^{2} + 2} d x

\frac{1}{6} ln 32 x + 1 + 18 x^{2} + C

Solution steps

\int \frac{1}{36 x ^{2} + 2} d x

Apply Trigonometric Substitution

: \int \frac{sec ( u )}{6} d u

= \int \frac{sec ( u )}{6} d u

Take the constant out:

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

= \frac{1}{6} \cdot \int sec (u) d u

Use the common integral:

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

= \frac{1}{6} ln ∣ tan (u) + sec (u) ∣

Substitute back

u = arctan (\frac{6}{2} x)

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

Simplify

\frac{1}{6} ln tan (arctan (\frac{6}{2} x)) + sec (arctan (\frac{6}{2} x)) : \frac{1}{6} ln 32 x + 1 + 18 x^{2}

= \frac{1}{6} ln 32 x + 1 + 18 x^{2}

Add a constant to the solution

= \frac{1}{6} ln 32 x + 1 + 18 x^{2} + C

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What is the integral of 1/(sqrt(36x^2+2)) ?
The integral of 1/(sqrt(36x^2+2)) is 1/6 ln|3sqrt(2)x+sqrt(1+18x^2)|+C