What is the integral from 0 to 9 of 1/(sqrt(81+t^2)) ?

The integral from 0 to 9 of 1/(sqrt(81+t^2)) is ln(1+sqrt(2))

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integral from 0 to 9 of 1/(sqrt(81+t^2))

\int_{0}^{9} \frac{1}{81 + t ^{2}} d t

ln (1 + 2)

Decimal

0.88137 \dots

Solution steps

\int_{0}^{9} \frac{1}{81 + 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)

x^{8}

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

\frac{1}{s ^{2} + 4 s + 20}

y^{^{'}} = 1 0^{- 21} \cdot y^{3}

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

What is the integral from 0 to 9 of 1/(sqrt(81+t^2)) ?
The integral from 0 to 9 of 1/(sqrt(81+t^2)) is ln(1+sqrt(2))