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

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

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

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

ln (1 + 2)

Decimal Notation

0.88137 \dots

Solution steps

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

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)

(34) (3 - 4)

\frac{d}{d x} (x^{2} + x y + y^{3})

cos (x)

\int_{1}^{3} x + 2 d x

3 y^{^{'}} + 12 y = 4

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