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

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

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

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

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

Solution steps

\int \frac{1}{16 + x ^{2}} 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{\partial}{\partial y} (8 x y - \frac{1}{4} (x + y)^{4})

x \to x lim (sin (x))

y = x^{2}, y = 9 x, y = - 1

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

\int \frac{4 x ^{3} - 2 x + 1}{x} d x

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