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

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

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integral from 0 to 1 of sqrt(3+9x^2)

\int_{0}^{1} 3 + 9 x^{2} d x

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

Decimal Notation

2.39052 \dots

Solution steps

\int_{0}^{1} 3 + 9 x^{2} d x

Apply Trigonometric Substitution

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

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

Apply Integral Reduction

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

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

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

Simplify

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

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

Compute the boundaries:

3

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

\frac{d}{d x} (\frac{e ^{x}}{9 x})

\int_{0}^{1} (2 x - 3) (5 x + 1) d x

\int \frac{1}{2} sin (x) d x

x \to 8 + lim (lo g_{5} (x - 8))

\int_{0}^{2} \frac{x}{x + 1} d x

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