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

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

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

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

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

Solution steps

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

Take the constant out:

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

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

Apply Trigonometric Substitution

: \int sec (u) d u

= - \int sec (u) d u

Use the common integral:

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

= - ln ∣ tan (u) + sec (u) ∣

Substitute back

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

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

Simplify

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

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

Add a constant to the solution

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

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

\frac{s}{9 s + 81}

\frac{d ^{2}}{d x ^{2}} (6 sec (x))

\int_{0}^{π} sin (x) cos (x) d x

x \to 0 lim (\frac{1}{x} \cdot sin (\frac{1}{x}))

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