What is the integral from 0 to 2 of sqrt(1+5t^2) ?

The integral from 0 to 2 of sqrt(1+5t^2) is 1/(sqrt(5))(sqrt(105)+1/2 ln(2sqrt(5)+sqrt(21)))

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integral from 0 to 2 of sqrt(1+5t^2)

\int_{0}^{2} 1 + 5 t^{2} d t

\frac{1}{5} (105 + \frac{1}{2} ln (25 + 21))

Decimal Notation

5.07524 \dots

Solution steps

\int_{0}^{2} 1 + 5 t^{2} d t

Apply Trigonometric Substitution

: \int_{0}^{a r c t a n (5 \cdot 2)} \frac{1}{5} sec^{3} (u) d u

= \int_{0}^{a r c t a n (5 \cdot 2)} \frac{1}{5} 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}{5} \cdot \int_{0}^{a r c t a n (5 \cdot 2)} sec^{3} (u) d u

Apply Integral Reduction

= \frac{1}{5} [\frac{sec ^{2} ( u ) sin ( u )}{2}]_{0}^{a r c t a n (5 \cdot 2)} + \frac{1}{2} \cdot \int_{0}^{a r c t a n (5 \cdot 2)} sec (u) d u

\int_{0}^{a r c t a n (5 \cdot 2)} sec (u) d u = ln (25 + 21)

= \frac{1}{5} [\frac{sec ^{2} ( u ) sin ( u )}{2}]_{0}^{a r c t a n (5 \cdot 2)} + \frac{1}{2} ln (25 + 21)

Simplify

\frac{1}{5} [\frac{sec ^{2} ( u ) sin ( u )}{2}]_{0}^{a r c t a n (5 \cdot 2)} + \frac{1}{2} ln (25 + 21) : \frac{1}{5} ([\frac{1}{2} sec (u) tan (u)]_{0}^{a r c t a n (25)} + \frac{1}{2} ln (25 + 21))

= \frac{1}{5} ([\frac{1}{2} sec (u) tan (u)]_{0}^{a r c t a n (25)} + \frac{1}{2} ln (25 + 21))

Compute the boundaries:

105

= \frac{1}{5} (105 + \frac{1}{2} ln (25 + 21))

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What is the integral from 0 to 2 of sqrt(1+5t^2) ?
The integral from 0 to 2 of sqrt(1+5t^2) is 1/(sqrt(5))(sqrt(105)+1/2 ln(2sqrt(5)+sqrt(21)))