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

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

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

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

ln (1 + 2)

Decimal Notation

0.88137 \dots

Solution steps

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

Apply Trigonometric Substitution

: \int_{0}^{\frac{π}{4}} sec (u) d u

= \int_{0}^{\frac{π}{4}} sec (u) d u

Use the common integral:

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

= [ln ∣ tan (u) + sec (u) ∣]_{0}^{\frac{π}{4}}

Compute the boundaries:

ln (1 + 2)

= ln (1 + 2)

\frac{\partial}{\partial x} (x^{4} y^{3} + 8 x^{2} y)

\frac{d}{d x} (\frac{x - 5}{5})

81 - y^{2}

y^{^{'}} = 6 - 3 y

x \to 0 lim (\frac{tan ( 12 x )}{x})

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