What is the integral from 0 to 6 of 1/(sqrt(36+x^2)) ?

The integral from 0 to 6 of 1/(sqrt(36+x^2)) is ln(1+sqrt(2))

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integral from 0 to 6 of 1/(sqrt(36+x^2))

\int_{0}^{6} \frac{1}{36 + x ^{2}} d x

ln (1 + 2)

Decimal Notation

0.88137 \dots

Solution steps

\int_{0}^{6} \frac{1}{36 + 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{\partial}{\partial x} (e^{3 y})

\int \frac{z}{z ^{4} + 4} d z

y^{^{''}} + 4 y^{^{'}} + 3 = 0

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

cos (x), 0

What is the integral from 0 to 6 of 1/(sqrt(36+x^2)) ?
The integral from 0 to 6 of 1/(sqrt(36+x^2)) is ln(1+sqrt(2))