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

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

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

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

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

Solution steps

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

Take the constant out:

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

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

Apply Trigonometric Substitution

: \int sec (u) d u

= - 2 \cdot \int sec (u) d u

Use the common integral:

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

= - 2 ln ∣ tan (u) + sec (u) ∣

Substitute back

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

= - 2 ln tan (arctan (\frac{1}{4} x)) + sec (arctan (\frac{1}{4} x))

Simplify

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

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

Add a constant to the solution

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

e^{\frac{y}{x}} \frac{( - y d x + x d y )}{x ^{2}} = 0

f (x) = \frac{8 x}{x ^{2} - 1}, (3, 3)

\frac{d}{d x} (3 x^{2} - 5 x + 4)

x \to 3 + lim (x^{2} - 9)

\frac{d}{d x} (\frac{x ^{2}}{x ^{4} + 16})

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