What is the integral from 0 to 4 of 1/(sqrt(9+x^2)) ?

The integral from 0 to 4 of 1/(sqrt(9+x^2)) is ln(3)

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integral from 0 to 4 of 1/(sqrt(9+x^2))

\int_{0}^{4} \frac{1}{9 + x ^{2}} d x

ln (3)

Decimal Notation

1.09861 \dots

Solution steps

\int_{0}^{4} \frac{1}{9 + x ^{2}} d x

Apply Trigonometric Substitution

: \int_{0}^{a r c t a n (\frac{4}{3})} sec (u) d u

= \int_{0}^{a r c t a n (\frac{4}{3})} sec (u) d u

Use the common integral:

\int sec (u) d u = ln ∣ tan (u) + sec (u) ∣

= [ln ∣ tan (u) + sec (u) ∣]_{0}^{a r c t a n (\frac{4}{3})}

Compute the boundaries:

ln (3)

= ln (3)

\frac{\partial}{\partial x} (- 6 x + 12)

y^{2} = 4 x, y = 4 x - 2

y = \frac{1}{x}, x = a

\frac{a}{y ^{10}}

f (x) = 4 x^{\frac{1}{2}}

What is the integral from 0 to 4 of 1/(sqrt(9+x^2)) ?
The integral from 0 to 4 of 1/(sqrt(9+x^2)) is ln(3)