What is the integral from 0 to 3 of 1/(sqrt(x^2+9)) ?

The integral from 0 to 3 of 1/(sqrt(x^2+9)) is ln(1+sqrt(2))

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integral from 0 to 3 of 1/(sqrt(x^2+9))

\int_{0}^{3} \frac{1}{x ^{2} + 9} d x

ln (1 + 2)

Decimal

0.88137 \dots

Solution steps

\int_{0}^{3} \frac{1}{x ^{2} + 9} 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)

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

\frac{d}{d x} (\frac{x}{x ^{2} + 6})

\frac{8 t}{24} - t^{5}

\frac{\partial}{\partial x} (sin^{2} (x - 5 y))

\frac{d}{d h} (\frac{π h ^{2}}{3} (36 - h))

What is the integral from 0 to 3 of 1/(sqrt(x^2+9)) ?
The integral from 0 to 3 of 1/(sqrt(x^2+9)) is ln(1+sqrt(2))