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

The integral from 0 to 2 of sqrt(1+3x^2) is 1/(sqrt(3))(sqrt(39)+1/2 ln(2sqrt(3)+sqrt(13)))

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

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

\frac{1}{3} (39 + \frac{1}{2} ln (23 + 13))

Decimal Notation

4.17014 \dots

Solution steps

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

Apply Trigonometric Substitution

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

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

Take the constant out:

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

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

Apply Integral Reduction

= \frac{1}{3} [\frac{sec ^{2} ( u ) sin ( u )}{2}]_{0}^{a r c t a n (3 \cdot 2)} + \frac{1}{2} \cdot \int_{0}^{a r c t a n (3 \cdot 2)} sec (u) d u

\int_{0}^{a r c t a n (3 \cdot 2)} sec (u) d u = ln (23 + 13)

= \frac{1}{3} [\frac{sec ^{2} ( u ) sin ( u )}{2}]_{0}^{a r c t a n (3 \cdot 2)} + \frac{1}{2} ln (23 + 13)

Simplify

\frac{1}{3} [\frac{sec ^{2} ( u ) sin ( u )}{2}]_{0}^{a r c t a n (3 \cdot 2)} + \frac{1}{2} ln (23 + 13) : \frac{1}{3} ([\frac{1}{2} sec (u) tan (u)]_{0}^{a r c t a n (23)} + \frac{1}{2} ln (23 + 13))

= \frac{1}{3} ([\frac{1}{2} sec (u) tan (u)]_{0}^{a r c t a n (23)} + \frac{1}{2} ln (23 + 13))

Compute the boundaries:

39

= \frac{1}{3} (39 + \frac{1}{2} ln (23 + 13))

6 x^{- 3}, at x = 5

\frac{x - y}{x + y}

\frac{d}{d x} (\frac{x ^{2}}{x})

\int_{0}^{1} 2 x^{2} - 1 d x

\frac{d x}{d t} = 2 x (5 - x)

What is the integral from 0 to 2 of sqrt(1+3x^2) ?
The integral from 0 to 2 of sqrt(1+3x^2) is 1/(sqrt(3))(sqrt(39)+1/2 ln(2sqrt(3)+sqrt(13)))