What is the integral of 3/(sqrt(x^2+16)) ?

The integral of 3/(sqrt(x^2+16)) is 3ln(1/4 |x+sqrt(16+x^2)|)+C

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integral of 3/(sqrt(x^2+16))

\int \frac{3}{x ^{2} + 16} d x

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

Solution steps

\int \frac{3}{x ^{2} + 16} d x

Take the constant out:

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

= 3 \cdot \int \frac{1}{x ^{2} + 16} d x

Apply Trigonometric Substitution

: \int sec (u) d u

= 3 \cdot \int sec (u) d u

Use the common integral:

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

= 3 ln ∣ tan (u) + sec (u) ∣

Substitute back

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

= 3 ln tan (arctan (\frac{1}{4} x)) + sec (arctan (\frac{1}{4} x))

Simplify

3 ln tan (arctan (\frac{1}{4} x)) + sec (arctan (\frac{1}{4} x)) : 3 ln (\frac{1}{4} x + 16 + x^{2})

= 3 ln (\frac{1}{4} x + 16 + x^{2})

Add a constant to the solution

= 3 ln (\frac{1}{4} x + 16 + x^{2}) + C

\int 5 x^{- 4} d x

f (x) = \frac{3}{x}, at x = 6

\frac{d}{d x} (5 ln (cos (x)))

x \to 0 + lim (\frac{1}{6 x})

x \to 0 lim (2 (\frac{a ^{x} - 1}{x}))

What is the integral of 3/(sqrt(x^2+16)) ?
The integral of 3/(sqrt(x^2+16)) is 3ln(1/4 |x+sqrt(16+x^2)|)+C