What is the integral from 0 to 3 of sqrt(x^2+9) ?

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

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integral from 0 to 3 of sqrt(x^2+9)

\int_{0}^{3} x^{2} + 9 d x

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

Decimal Notation

10.33014 \dots

Solution steps

\int_{0}^{3} x^{2} + 9 d x

Apply Trigonometric Substitution

: \int_{0}^{\frac{π}{4}} 9 sec^{3} (u) d u

= \int_{0}^{\frac{π}{4}} 9 sec^{3} (u) d u

Take the constant out:

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

= 9 \cdot \int_{0}^{\frac{π}{4}} sec^{3} (u) d u

Apply Integral Reduction

= 9 ([\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)

= 9 ([\frac{sec ^{2} ( u ) sin ( u )}{2}]_{0}^{\frac{π}{4}} + \frac{1}{2} ln (1 + 2))

Simplify

9 ([\frac{sec ^{2} ( u ) sin ( u )}{2}]_{0}^{\frac{π}{4}} + \frac{1}{2} ln (1 + 2)) : 9 ([\frac{1}{2} sec (u) tan (u)]_{0}^{\frac{π}{4}} + \frac{1}{2} ln (1 + 2))

= 9 ([\frac{1}{2} sec (u) tan (u)]_{0}^{\frac{π}{4}} + \frac{1}{2} ln (1 + 2))

Compute the boundaries:

\frac{1}{2}

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

Simplify

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

y = 2 x^{3} + 3 x^{2} - 12 x + 2

\int x^{2} - x - 6 d x

\frac{d}{d x} (4 sin (4 x))

\frac{\partial}{\partial y} (x y + y^{2})

y^{^{'}} = y (1000 - y)

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