What is the integral from 0 to 5 of 1/(sqrt(9+4x^2)) ?

The integral from 0 to 5 of 1/(sqrt(9+4x^2)) is 1/2 ln((10+sqrt(109))/3)

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integral from 0 to 5 of 1/(sqrt(9+4x^2))

\int_{0}^{5} \frac{1}{9 + 4 x ^{2}} d x

\frac{1}{2} ln (\frac{10 + 109}{3})

Decimal Notation

0.95944 \dots

Solution steps

\int_{0}^{5} \frac{1}{9 + 4 x ^{2}} d x

Apply Trigonometric Substitution

: \int_{0}^{a r c t a n (\frac{10}{3})} \frac{sec ( u )}{2} d u

= \int_{0}^{a r c t a n (\frac{10}{3})} \frac{sec ( u )}{2} d u

Take the constant out:

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

= \frac{1}{2} \cdot \int_{0}^{a r c t a n (\frac{10}{3})} sec (u) d u

Use the common integral:

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

= \frac{1}{2} [ln ∣ tan (u) + sec (u) ∣]_{0}^{a r c t a n (\frac{10}{3})}

Compute the boundaries:

ln (\frac{10}{3} + \frac{109}{3})

= \frac{1}{2} ln (\frac{10}{3} + \frac{109}{3})

Simplify

= \frac{1}{2} ln (\frac{10 + 109}{3})

x \to 6 lim (3 x)

\int \frac{cos ( x )}{1 - sin ^{2} ( x )} d x

\int (\frac{1}{x ^{8}}) d x

\frac{d}{d x} (25 - x)

What is the integral from 0 to 5 of 1/(sqrt(9+4x^2)) ?
The integral from 0 to 5 of 1/(sqrt(9+4x^2)) is 1/2 ln((10+sqrt(109))/3)