What is the integral from 0 to 7 of 1/(sqrt(49+x^2)) ?

The integral from 0 to 7 of 1/(sqrt(49+x^2)) is ln(1+sqrt(2))

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integral from 0 to 7 of 1/(sqrt(49+x^2))

\int_{0}^{7} \frac{1}{49 + x ^{2}} d x

ln (1 + 2)

Decimal Notation

0.88137 \dots

Solution steps

\int_{0}^{7} \frac{1}{49 + 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{1}{4 - y ^{2}} d y

y^{^{'}} = sin (y)

\int \frac{8 t ^{3} - 3 t ^{2} + 32 t}{t ^{2} + 4} d t

\frac{\partial}{\partial y} (x^{7 y})

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

What is the integral from 0 to 7 of 1/(sqrt(49+x^2)) ?
The integral from 0 to 7 of 1/(sqrt(49+x^2)) is ln(1+sqrt(2))