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

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

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

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

ln (1 + 2)

Solution steps

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

y = sin (x) cos (x)

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

y^{^{'}} - 6 y = 0

\frac{\partial}{\partial x} (x^{0.2} \cdot y^{0.8})

\int 4 cos (x - 1) d x

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