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

The integral from 0 to sqrt(3 of)sqrt(x^2+9) is 9(1/3+1/4 ln(3))

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

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

9 (\frac{1}{3} + \frac{1}{4} ln (3))

Decimal Notation

5.47187 \dots

Solution steps

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

Apply Trigonometric Substitution

: \int_{0}^{\frac{π}{6}} 9 sec^{3} (u) d u

= \int_{0}^{\frac{π}{6}} 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}{6}} sec^{3} (u) d u

Apply Integral Reduction

= 9 ([\frac{sec ^{2} ( u ) sin ( u )}{2}]_{0}^{\frac{π}{6}} + \frac{1}{2} \cdot \int_{0}^{\frac{π}{6}} sec (u) d u)

\int_{0}^{\frac{π}{6}} sec (u) d u = \frac{1}{2} ln (3)

= 9 ([\frac{sec ^{2} ( u ) sin ( u )}{2}]_{0}^{\frac{π}{6}} + \frac{1}{2} \cdot \frac{1}{2} ln (3))

Simplify

9 ([\frac{sec ^{2} ( u ) sin ( u )}{2}]_{0}^{\frac{π}{6}} + \frac{1}{2} \cdot \frac{1}{2} ln (3)) : 9 ([\frac{1}{2} sec (u) tan (u)]_{0}^{\frac{π}{6}} + \frac{1}{4} ln (3))

= 9 ([\frac{1}{2} sec (u) tan (u)]_{0}^{\frac{π}{6}} + \frac{1}{4} ln (3))

Compute the boundaries:

\frac{1}{3}

= 9 (\frac{1}{3} + \frac{1}{4} ln (3))

\frac{d}{d x} (\frac{1}{4 x ^{2}})

\frac{d}{d x} (\frac{x ^{2} - 3}{x ^{2} + 1})

\int_{1}^{5} t^{3} ln (4 t) d t

\int \frac{x - 1}{x ^{2} - 2 x + 3} d x

3 x^{- \frac{1}{2}}

What is the integral from 0 to sqrt(3 of)sqrt(x^2+9) ?
The integral from 0 to sqrt(3 of)sqrt(x^2+9) is 9(1/3+1/4 ln(3))