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))

\frac{s ^{2} + 8}{( s + 2 ) ^{4}}

n = - 4 \sum \infty (- \frac{4}{11})^{n}

\frac{\partial}{\partial z} (x + z)

\frac{2}{s \cdot ( s + 2 )}

\int_{2}^{7} t^{4} ln (2 t) d t

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)))