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

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

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

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

ln (3)

Decimal Notation

1.09861 \dots

Solution steps

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

3 x - 2 x

\int e^{x} \cdot sin (x) d x

\frac{d}{d x} (\frac{x ^{2} + 1}{x ^{3} - 1})

f (x) = 17 e^{e} - e x^{e}

e^{- t} t sin (2 t)

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