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

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

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

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

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

Decimal

1.27127 \dots

Solution steps

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

Apply Trigonometric Substitution

: \int_{0}^{a r c t a n (2)} \frac{1}{2} sec^{3} (u) d u

= \int_{0}^{a r c t a n (2)} \frac{1}{2} sec^{3} (u) d u

Take the constant out:

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

= \frac{1}{2} \cdot \int_{0}^{a r c t a n (2)} sec^{3} (u) d u

Apply Integral Reduction

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

\int_{0}^{a r c t a n (2)} sec (u) d u = ln (2 + 3)

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

Simplify

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

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

Compute the boundaries:

\frac{3}{2}

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

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