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

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

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

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

ln (1 + 2)

Decimal Notation

0.88137 \dots

Solution steps

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

\int_{- 2}^{0} x^{3} - 4 x d x

\frac{\partial}{\partial x} (60 e^{x^{6} y} x^{5} y)

\int_{4}^{5} \frac{x}{x - 4} d x

\frac{\partial}{\partial x} ((y^{2} - 2)^{3})

\frac{1}{3 s + 5}

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