What is the integral of-8/(sqrt(x^2+16)) ?

The integral of-8/(sqrt(x^2+16)) is -8ln(1/4 |x+sqrt(16+x^2)|)+C

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integral of-8/(sqrt(x^2+16))

\int - \frac{8}{x ^{2} + 16} d x

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

Solution steps

\int - \frac{8}{x ^{2} + 16} d x

Take the constant out:

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

= - 8 \cdot \int \frac{1}{x ^{2} + 16} d x

Apply Trigonometric Substitution

: \int sec (u) d u

= - 8 \cdot \int sec (u) d u

Use the common integral:

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

= - 8 ln ∣ tan (u) + sec (u) ∣

Substitute back

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

= - 8 ln tan (arctan (\frac{1}{4} x)) + sec (arctan (\frac{1}{4} x))

Simplify

- 8 ln tan (arctan (\frac{1}{4} x)) + sec (arctan (\frac{1}{4} x)) : - 8 ln (\frac{1}{4} x + 16 + x^{2})

= - 8 ln (\frac{1}{4} x + 16 + x^{2})

Add a constant to the solution

= - 8 ln (\frac{1}{4} x + 16 + x^{2}) + C

f (x) = 8 e^{- 0.5 x}, at x = 7

\frac{\partial}{\partial x} (4 x sin (x y))

\frac{d}{d x} (- \frac{2}{( x - 2 ) ^{2}})

x \to 2 lim ((\frac{x + 3}{2 x + 1})^{\frac{2}{x - 2}})

\frac{d}{d x} (x arccos (3 x))

What is the integral of-8/(sqrt(x^2+16)) ?
The integral of-8/(sqrt(x^2+16)) is -8ln(1/4 |x+sqrt(16+x^2)|)+C