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

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

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integral of 4/(sqrt(x^2-16))

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

4 ln (\frac{1}{4} x^{2} - 16 + x) + C

Solution steps

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

Take the constant out:

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

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

Apply Trigonometric Substitution

: \int sec (u) d u

= 4 \cdot \int sec (u) d u

Use the common integral:

\int sec (u) d u = ln ∣ tan (u) + sec (u) ∣

= 4 ln ∣ tan (u) + sec (u) ∣

Substitute back

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

= 4 ln tan (arcsec (\frac{1}{4} x)) + sec (arcsec (\frac{1}{4} x))

Simplify

4 ln tan (arcsec (\frac{1}{4} x)) + sec (arcsec (\frac{1}{4} x)) : 4 ln (\frac{1}{4} x^{2} - 16 + x)

= 4 ln (\frac{1}{4} x^{2} - 16 + x)

Add a constant to the solution

= 4 ln (\frac{1}{4} x^{2} - 16 + x) + C

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

\frac{\partial}{\partial x} (x \cdot \frac{e ^{y}}{x + y})

n = 8 \sum \infty \frac{n}{n ln ( n )}

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

f (x) = \frac{( x ^{2} - 1 )}{( x ^{2} + x + 1 )}

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