What is the integral of 1/(4xsqrt(x^2-16)) ?

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

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integral of 1/(4xsqrt(x^2-16))

\int \frac{1}{4 x x ^{2} - 16} d x

\frac{1}{16} arcsec (\frac{1}{4} x) + C

Solution steps

\int \frac{1}{4 x x ^{2} - 16} d x

Take the constant out:

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

= \frac{1}{4} \cdot \int \frac{1}{x x ^{2} - 16} d x

Apply Trigonometric Substitution

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

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

Integral of a constant:

\int a d x = a x

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

Substitute back

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

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

Simplify

\frac{1}{4} \cdot \frac{1}{4} \cdot 1 \cdot arcsec (\frac{1}{4} x) : \frac{1}{16} arcsec (\frac{1}{4} x)

= \frac{1}{16} arcsec (\frac{1}{4} x)

Add a constant to the solution

= \frac{1}{16} arcsec (\frac{1}{4} x) + C

\frac{d}{d x} (1 - cos (5 x))

x \to 1 lim (\frac{x + 3}{x + 1})

- 6 t^{2} δ (t - 5)

5 - 5 x^{3}

\frac{d}{d x} (\frac{2 x}{7 - cot ( x )})

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