What is the integral of 1/(xsqrt(16x^2-36)) ?

The integral of 1/(xsqrt(16x^2-36)) is 1/6 arcsec((2x)/3)+C

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integral of 1/(xsqrt(16x^2-36))

\int \frac{1}{x 16 x ^{2} - 36} d x

\frac{1}{6} arcsec (\frac{2 x}{3}) + C

Solution steps

\int \frac{1}{x 16 x ^{2} - 36} d x

Apply Trigonometric Substitution

: \int \frac{1}{6} d u

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

Integral of a constant:

\int a d x = a x

= \frac{1}{6} \cdot 1 \cdot u

Substitute back

u = arcsec (\frac{4}{6} x)

= \frac{1}{6} \cdot 1 \cdot arcsec (\frac{4}{6} x)

Simplify

\frac{1}{6} \cdot 1 \cdot arcsec (\frac{4}{6} x) : \frac{1}{6} arcsec (\frac{2 x}{3})

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

Add a constant to the solution

= \frac{1}{6} arcsec (\frac{2 x}{3}) + C

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What is the integral of 1/(xsqrt(16x^2-36)) ?
The integral of 1/(xsqrt(16x^2-36)) is 1/6 arcsec((2x)/3)+C