What is the integral of 1/(2xsqrt(16x^2-8)) ?

The integral of 1/(2xsqrt(16x^2-8)) is 1/(4sqrt(2))arcsec(sqrt(2)x)+C

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integral of 1/(2xsqrt(16x^2-8))

\int \frac{1}{2 x 16 x ^{2} - 8} d x

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

Solution steps

\int \frac{1}{2 x 16 x ^{2} - 8} d x

Take the constant out:

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

= \frac{1}{2} \cdot \int \frac{1}{x 16 x ^{2} - 8} d x

Apply Trigonometric Substitution

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

Integral of a constant:

\int a d x = a x

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

Substitute back

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

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

Simplify

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

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

Add a constant to the solution

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

What is the integral of 1/(2xsqrt(16x^2-8)) ?
The integral of 1/(2xsqrt(16x^2-8)) is 1/(4sqrt(2))arcsec(sqrt(2)x)+C