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

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

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

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

4 ln (3)

Decimal Notation

4.39444 \dots

Solution steps

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

Take the constant out:

\int a \cdot f (x) d x = a \cdot \int f (x) d x

= 4 \cdot \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

= 4 \cdot \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) ∣

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

Compute the boundaries:

ln (3)

= 4 ln (3)

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