What is the integral of 1/(8xsqrt(x^2-64)) ?

The integral of 1/(8xsqrt(x^2-64)) is 1/64 arcsec(1/8 x)+C

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integral of 1/(8xsqrt(x^2-64))

\int \frac{1}{8 x x ^{2} - 64} d x

\frac{1}{64} arcsec (\frac{1}{8} x) + C

Solution steps

\int \frac{1}{8 x x ^{2} - 64} d x

Take the constant out:

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

= \frac{1}{8} \cdot \int \frac{1}{x x ^{2} - 64} d x

Apply Trigonometric Substitution

: \int \frac{1}{8} d u

= \frac{1}{8} \cdot \int \frac{1}{8} d u

Take the constant out:

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

= \frac{1}{8} \cdot \frac{1}{8} \cdot \int 1 d u

Integral of a constant:

\int a d x = a x

= \frac{1}{8} \cdot \frac{1}{8} \cdot 1 \cdot u

Substitute back

u = arcsec (\frac{1}{8} x)

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

Simplify

\frac{1}{8} \cdot \frac{1}{8} \cdot 1 \cdot arcsec (\frac{1}{8} x) : \frac{1}{64} arcsec (\frac{1}{8} x)

= \frac{1}{64} arcsec (\frac{1}{8} x)

Add a constant to the solution

= \frac{1}{64} arcsec (\frac{1}{8} x) + C

y^{^{'}} = \frac{ln ( 2 )}{7} y - 20000, y (0) = 200000

\frac{d}{d x} (x (x^{4} + 9)^{3})

\int ln^{2} (\frac{1}{2} x) d x

\frac{( 12 s + 8 )}{( s - 1 ) ^{5}}

(1 - x) y^{^{'}} = y^{2}

What is the integral of 1/(8xsqrt(x^2-64)) ?
The integral of 1/(8xsqrt(x^2-64)) is 1/64 arcsec(1/8 x)+C