What is the integral from 0 to 5 of 1/(sqrt(25+x^2)) ?

The integral from 0 to 5 of 1/(sqrt(25+x^2)) is ln(1+sqrt(2))

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integral from 0 to 5 of 1/(sqrt(25+x^2))

\int_{0}^{5} \frac{1}{25 + x ^{2}} d x

ln (1 + 2)

Decimal Notation

0.88137 \dots

Solution steps

\int_{0}^{5} \frac{1}{25 + 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)

\int \frac{c}{x ^{2}} d x

f (x) = 7 x^{15}

\int_{0}^{5} \frac{1}{x ^{\frac{18}{19}}} d x

f (x) = \frac{3}{4 x + 1}, at x = - 1

\frac{d}{d x} (\frac{e ^{a r c t a n (x)}}{1 + x ^{2}})

What is the integral from 0 to 5 of 1/(sqrt(25+x^2)) ?
The integral from 0 to 5 of 1/(sqrt(25+x^2)) is ln(1+sqrt(2))