What is the integral from 0 to 9 of sqrt(81+x^2) ?

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

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integral from 0 to 9 of sqrt(81+x^2)

\int_{0}^{9} 81 + x^{2} d x

\frac{81 ( 2 + ln ( 1 + 2 ) )}{2}

Decimal

92.97127 \dots

Solution steps

\int_{0}^{9} 81 + x^{2} d x

Apply Trigonometric Substitution

: \int_{0}^{\frac{π}{4}} 81 sec^{3} (u) d u

= \int_{0}^{\frac{π}{4}} 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}^{\frac{π}{4}} sec^{3} (u) d u

Apply Integral Reduction

= 81 ([\frac{sec ^{2} ( u ) sin ( u )}{2}]_{0}^{\frac{π}{4}} + \frac{1}{2} \cdot \int_{0}^{\frac{π}{4}} sec (u) d u)

\int_{0}^{\frac{π}{4}} sec (u) d u = ln (1 + 2)

= 81 ([\frac{sec ^{2} ( u ) sin ( u )}{2}]_{0}^{\frac{π}{4}} + \frac{1}{2} ln (1 + 2))

Simplify

81 ([\frac{sec ^{2} ( u ) sin ( u )}{2}]_{0}^{\frac{π}{4}} + \frac{1}{2} ln (1 + 2)) : 81 ([\frac{1}{2} sec (u) tan (u)]_{0}^{\frac{π}{4}} + \frac{1}{2} ln (1 + 2))

= 81 ([\frac{1}{2} sec (u) tan (u)]_{0}^{\frac{π}{4}} + \frac{1}{2} ln (1 + 2))

Compute the boundaries:

\frac{1}{2}

= 81 (\frac{1}{2} + \frac{1}{2} ln (1 + 2))

Simplify

= \frac{81 ( 2 + ln ( 1 + 2 ) )}{2}

\int \frac{1}{441 + x ^{2}} d x

\int - \frac{x ^{3}}{4} d x

\int (x^{2} + x) e^{x} d x

x \to 0 lim ((1 - 9 x)^{\frac{1}{x}})

\frac{d}{d x} (4 9 x^{2} + 5)

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