What is the integral of 1/((x*sqrt(x^2-16))) ?

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

Popular >

integral of 1/((x*sqrt(x^2-16)))

\int \frac{1}{( x \cdot x ^{2} - 16 )} d x

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

Solution steps

\int \frac{1}{x 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{1}{4} x)

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

Simplify

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

Add a constant to the solution

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

y = e^{- x} cos (x)

\int \frac{1}{x ^{3} - 4 x ^{2}} d x

\frac{\partial}{\partial x} (\frac{9 x}{7 y + 8 z})

\frac{x d y}{d x} + 6 y = x^{3} - x

\int 8 x e^{2 x} d x

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