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

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

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

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

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

Decimal

1.52450 \dots

Solution steps

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

Apply Trigonometric Substitution

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

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

Take the constant out:

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

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

Apply Integral Reduction

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

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

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

Simplify

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

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

Compute the boundaries:

\frac{3}{4}

= 2 \cdot \frac{3}{4} + ln (\frac{1}{2} + \frac{3}{2})

Simplify

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

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