What is the integral from 0 to 3 of 1/(sqrt(64+x^2)) ?

The integral from 0 to 3 of 1/(sqrt(64+x^2)) is ln(3/8+(sqrt(73))/8)

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integral from 0 to 3 of 1/(sqrt(64+x^2))

\int_{0}^{3} \frac{1}{64 + x ^{2}} d x

ln (\frac{3}{8} + \frac{73}{8})

Decimal

0.36672 \dots

Solution steps

\int_{0}^{3} \frac{1}{64 + x ^{2}} d x

Apply Trigonometric Substitution

: \int_{0}^{a r c t a n (\frac{3}{8})} sec (u) d u

= \int_{0}^{a r c t a n (\frac{3}{8})} sec (u) d u

Use the common integral:

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

= [ln ∣ tan (u) + sec (u) ∣]_{0}^{a r c t a n (\frac{3}{8})}

Compute the boundaries:

ln (\frac{3}{8} + \frac{73}{8})

= ln (\frac{3}{8} + \frac{73}{8})

arcsin (\frac{x}{3})

\frac{d ^{4}}{d x ^{4}} (cos (3 x))

n = 0 \sum \infty - 1

\int_{- 3}^{3} x 9 - x^{2} d x

\int tan^{4} (θ) sec^{6} (θ) d θ

What is the integral from 0 to 3 of 1/(sqrt(64+x^2)) ?
The integral from 0 to 3 of 1/(sqrt(64+x^2)) is ln(3/8+(sqrt(73))/8)