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

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

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

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

ln (1 + 2)

Decimal

0.88137 \dots

Solution steps

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

n = 0 \sum \infty (0.8)^{n}

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

\frac{d}{d x} ((e^{(x^{2})} + 4)^{2} (4 x)^{3 x})

\frac{d}{d t} (t sin (π) t)

\frac{d}{d x} (9 x - x^{2})

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