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

The integral of 1/(x^2sqrt(x^2-64)) is (sqrt(x^2-64))/(64x)+C

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integral of 1/(x^2sqrt(x^2-64))

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

\frac{x ^{2} - 64}{64 x} + C

Solution steps

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

Apply Trigonometric Substitution

: \int \frac{1}{64 sec ( u )} d u

= \int \frac{1}{64 sec ( u )} d u

Take the constant out:

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

= \frac{1}{64} \cdot \int \frac{1}{sec ( u )} d u

Rewrite using trig identities

= \frac{1}{64} \cdot \int cos (u) d u

Use the common integral:

\int cos (u) d u = sin (u)

= \frac{1}{64} sin (u)

Substitute back

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

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

Simplify

\frac{1}{64} sin (arcsec (\frac{1}{8} x)) : \frac{x ^{2} - 64}{64 x}

= \frac{x ^{2} - 64}{64 x}

Add a constant to the solution

= \frac{x ^{2} - 64}{64 x} + C

\int \frac{1}{t ^{2} - 8 t + 25} d t

\int e^{2 x + 5} d x

\frac{d y}{d x} = \frac{2 x}{y ^{2}}

\int \frac{13}{( x - 3 ) ( x ^{2} + 4 )} d x

f (x) = \frac{2 ln ( x )}{x}

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