What is the integral from 0 to 2 of 1/(sqrt(16+x^2)) ?

The integral from 0 to 2 of 1/(sqrt(16+x^2)) is ln(1/2+(sqrt(5))/2)

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integral from 0 to 2 of 1/(sqrt(16+x^2))

\int_{0}^{2} \frac{1}{16 + x ^{2}} d x

ln (\frac{1}{2} + \frac{5}{2})

Solution steps

\int_{0}^{2} \frac{1}{16 + x ^{2}} d x

Apply Trigonometric Substitution

: \int_{0}^{a r c t a n (\frac{1}{2})} sec (u) d u

= \int_{0}^{a r c t a n (\frac{1}{2})} sec (u) d u

Use the common integral:

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

= [ln ∣ tan (u) + sec (u) ∣]_{0}^{a r c t a n (\frac{1}{2})}

Compute the boundaries:

ln (\frac{1}{2} + \frac{5}{2})

= ln (\frac{1}{2} + \frac{5}{2})

\int \frac{x}{x - 9} d x

4 x^{2} + 2 x - 1, at x = 0

x y^{2} y^{^{'}} = x^{3} + 2 y^{3}

\frac{30}{s ( s + 5 ) ( s + 4 )}

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

What is the integral from 0 to 2 of 1/(sqrt(16+x^2)) ?
The integral from 0 to 2 of 1/(sqrt(16+x^2)) is ln(1/2+(sqrt(5))/2)