What is the integral from 0 to 1 of 1/(sqrt(4+x^2)) ?

The integral from 0 to 1 of 1/(sqrt(4+x^2)) is ln(1/2+(sqrt(5))/2)

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integral from 0 to 1 of 1/(sqrt(4+x^2))

\int_{0}^{1} \frac{1}{4 + x ^{2}} d x

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

Decimal

0.48121 \dots

Solution steps

\int_{0}^{1} \frac{1}{4 + 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})

n = 0 \sum \infty \frac{1}{2 ^{n} + n}

\int \frac{x - ln ^{2} ( x )}{x} d x

\int (4 - 4 cos (x))^{2} d x

y = 4 x - x^{2}, at x = a

\frac{d}{d x} (2 x (4 x + 9)^{\frac{1}{2}} + 12)

What is the integral from 0 to 1 of 1/(sqrt(4+x^2)) ?
The integral from 0 to 1 of 1/(sqrt(4+x^2)) is ln(1/2+(sqrt(5))/2)