What is the integral of 4/(sqrt(x^2+16)) ?

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

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integral of 4/(sqrt(x^2+16))

\int \frac{4}{x ^{2} + 16} d x

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

Solution steps

\int \frac{4}{x ^{2} + 16} d x

Take the constant out:

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

= 4 \cdot \int \frac{1}{x ^{2} + 16} d x

Apply Trigonometric Substitution

: \int sec (u) d u

= 4 \cdot \int sec (u) d u

Use the common integral:

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

= 4 ln ∣ tan (u) + sec (u) ∣

Substitute back

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

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

Simplify

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

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

Add a constant to the solution

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

f (v) = (v^{- 3} + 6 v^{- 2})^{3}

f (x) = \frac{3}{x + 1}

\frac{\partial}{\partial z} (x y cos (x + yz))

\frac{d y}{d x} = 20 - 5 y

\int \frac{2 - 8 x}{1 - 4 x ^{2}} d x

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