What is the integral from 0 to 3 of 1/(sqrt(16+x^2)) ?

The integral from 0 to 3 of 1/(sqrt(16+x^2)) is ln(2)

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integral from 0 to 3 of 1/(sqrt(16+x^2))

\int_{0}^{3} \frac{1}{16 + x ^{2}} d x

ln (2)

Decimal

0.69314 \dots

Solution steps

\int_{0}^{3} \frac{1}{16 + x ^{2}} d x

Apply Trigonometric Substitution

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

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

Compute the boundaries:

ln (2)

= ln (2)

\int_{- 2}^{2} t^{2} d t

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

\int_{1}^{2} \frac{v ^{5} + 4 v ^{7}}{v ^{4}} d v

\int_{- \frac{π}{2}}^{\frac{π}{2}} 3 sin (x) d x

\int_{0}^{3} \frac{x ^{2}}{x ^{3} + 1} d x

What is the integral from 0 to 3 of 1/(sqrt(16+x^2)) ?
The integral from 0 to 3 of 1/(sqrt(16+x^2)) is ln(2)