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

The integral from 0 to 2 of sqrt(x^2+2) is sqrt(6)+ln(sqrt(2)+sqrt(3))

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

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

6 + ln (2 + 3)

Decimal Notation

3.59570 \dots

Solution steps

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

Apply Trigonometric Substitution

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

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

Apply Integral Reduction

= 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)

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

Simplify

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

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

Compute the boundaries:

\frac{3}{2}

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

Simplify

= 6 + ln (2 + 3)

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What is the integral from 0 to 2 of sqrt(x^2+2) ?
The integral from 0 to 2 of sqrt(x^2+2) is sqrt(6)+ln(sqrt(2)+sqrt(3))