What is the integral from 0 to 1 of sqrt(81x^2+729) ?

The integral from 0 to 1 of sqrt(81x^2+729) is (9(9ln((1+sqrt(10))/3)+sqrt(10)))/(2)

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integral from 0 to 1 of sqrt(81x^2+729)

\int_{0}^{1} 81 x^{2} + 729 d x

\frac{9 ( 9 ln ( \frac{1 + 10}{3} ) + 10 )}{2}

Decimal

27.49198 \dots

Solution steps

\int_{0}^{1} 81 x^{2} + 729 d x

Apply Trigonometric Substitution

: \int_{0}^{a r c t a n (\frac{1}{3})} 81 sec^{3} (u) d u

= \int_{0}^{a r c t a n (\frac{1}{3})} 81 sec^{3} (u) d u

Take the constant out:

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

= 81 \cdot \int_{0}^{a r c t a n (\frac{1}{3})} sec^{3} (u) d u

Apply Integral Reduction

= 81 ([\frac{sec ^{2} ( u ) sin ( u )}{2}]_{0}^{a r c t a n (\frac{1}{3})} + \frac{1}{2} \cdot \int_{0}^{a r c t a n (\frac{1}{3})} sec (u) d u)

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

= 81 ([\frac{sec ^{2} ( u ) sin ( u )}{2}]_{0}^{a r c t a n (\frac{1}{3})} + \frac{1}{2} ln (\frac{1}{3} + \frac{10}{3}))

Simplify

81 ([\frac{sec ^{2} ( u ) sin ( u )}{2}]_{0}^{a r c t a n (\frac{1}{3})} + \frac{1}{2} ln (\frac{1}{3} + \frac{10}{3})) : 81 ([\frac{1}{2} sec (u) tan (u)]_{0}^{a r c t a n (\frac{1}{3})} + \frac{1}{2} ln (\frac{1}{3} + \frac{10}{3}))

= 81 ([\frac{1}{2} sec (u) tan (u)]_{0}^{a r c t a n (\frac{1}{3})} + \frac{1}{2} ln (\frac{1}{3} + \frac{10}{3}))

Compute the boundaries:

\frac{5}{9 2}

= 81 (\frac{5}{9 2} + \frac{1}{2} ln (\frac{1}{3} + \frac{10}{3}))

Simplify

= \frac{9 ( 9 ln ( \frac{1 + 10}{3} ) + 10 )}{2}

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What is the integral from 0 to 1 of sqrt(81x^2+729) ?
The integral from 0 to 1 of sqrt(81x^2+729) is (9(9ln((1+sqrt(10))/3)+sqrt(10)))/(2)