What is the integral of 1/(xsqrt(64x^2-16)) ?

The integral of 1/(xsqrt(64x^2-16)) is 1/4 arcsec(2x)+C

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integral of 1/(xsqrt(64x^2-16))

\int \frac{1}{x 64 x ^{2} - 16} d x

\frac{1}{4} arcsec (2 x) + C

Solution steps

\int \frac{1}{x 64 x ^{2} - 16} d x

Apply Trigonometric Substitution

: \int \frac{1}{4} d u

= \int \frac{1}{4} d u

Take the constant out:

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

= \frac{1}{4} \cdot \int 1 d u

Integral of a constant:

\int a d x = a x

= \frac{1}{4} \cdot 1 \cdot u

Substitute back

u = arcsec (\frac{8}{4} x)

= \frac{1}{4} \cdot 1 \cdot arcsec (\frac{8}{4} x)

Simplify

\frac{1}{4} \cdot 1 \cdot arcsec (\frac{8}{4} x) : \frac{1}{4} arcsec (2 x)

= \frac{1}{4} arcsec (2 x)

Add a constant to the solution

= \frac{1}{4} arcsec (2 x) + C

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What is the integral of 1/(xsqrt(64x^2-16)) ?
The integral of 1/(xsqrt(64x^2-16)) is 1/4 arcsec(2x)+C