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

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

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integral from 0 to 8 of 1/(sqrt(64+x^2))

\int_{0}^{8} \frac{1}{64 + x ^{2}} d x

ln (1 + 2)

Decimal Notation

0.88137 \dots

Solution steps

\int_{0}^{8} \frac{1}{64 + 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)

\frac{d}{d x} (\frac{x sin ( 2 x )}{cos ( 3 x )})

8 t \frac{d y}{d t} + y = t^{5}

f (x) = x^{2} + 3 x

\int_{- π}^{0} sin (x) + x d x

\frac{d}{d x} (e^{6 x} cos (b x))

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