What is the integral of 1/(xsqrt(36x^2-4)) ?

The integral of 1/(xsqrt(36x^2-4)) is 1/2 arcsec(3x)+C

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integral of 1/(xsqrt(36x^2-4))

\int \frac{1}{x 36 x ^{2} - 4} d x

\frac{1}{2} arcsec (3 x) + C

Solution steps

\int \frac{1}{x 36 x ^{2} - 4} d x

Apply Trigonometric Substitution

: \int \frac{1}{2} d u

= \int \frac{1}{2} d u

Take the constant out:

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

= \frac{1}{2} \cdot \int 1 d u

Integral of a constant:

\int a d x = a x

= \frac{1}{2} \cdot 1 \cdot u

Substitute back

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

= \frac{1}{2} \cdot 1 \cdot arcsec (\frac{6}{2} x)

Simplify

\frac{1}{2} \cdot 1 \cdot arcsec (\frac{6}{2} x) : \frac{1}{2} arcsec (3 x)

= \frac{1}{2} arcsec (3 x)

Add a constant to the solution

= \frac{1}{2} arcsec (3 x) + C

\frac{d}{d x} (x^{2} + 3 x + 2)

6 \cdot 6^{t} + 2 \cdot 2 t

\int \frac{sin ( 3 x )}{csc ^{2} ( 4 x )} d x

\frac{d y}{d x} = \frac{1 - y}{1 - x}

\frac{\partial}{\partial v} (2 v sin (u))

What is the integral of 1/(xsqrt(36x^2-4)) ?
The integral of 1/(xsqrt(36x^2-4)) is 1/2 arcsec(3x)+C