What is the integral from 0 to 4 of 1/(sqrt(16+x^2)) ?

The integral from 0 to 4 of 1/(sqrt(16+x^2)) is ln(1+sqrt(2))

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integral from 0 to 4 of 1/(sqrt(16+x^2))

\int_{0}^{4} \frac{1}{16 + x ^{2}} d x

ln (1 + 2)

Decimal

0.88137 \dots

Solution steps

\int_{0}^{4} \frac{1}{16 + x ^{2}} d x

Apply Trigonometric Substitution

= \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} (2 cos (2 x) - 2 sin (2 x))

\frac{d}{d x} (ln 1 + x - x^{3})

\int \frac{1}{x 1 + x ^{n}} d x

x \to \infty lim ((a)^{x})

y^{^{'}} + 3 x^{2} y = sin (x) e^{- x^{3}}, y (0) = 1

What is the integral from 0 to 4 of 1/(sqrt(16+x^2)) ?
The integral from 0 to 4 of 1/(sqrt(16+x^2)) is ln(1+sqrt(2))