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

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

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

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

3 + \frac{1}{2 2} ln (22 + 3)

Decimal Notation

3.62322 \dots

Solution steps

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

Apply Trigonometric Substitution

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

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

\int_{0}^{a r c t a n (2 \cdot 2)} sec (u) d u = ln (22 + 3)

= \frac{1}{2} [\frac{sec ^{2} ( u ) sin ( u )}{2}]_{0}^{a r c t a n (2 \cdot 2)} + \frac{1}{2} ln (22 + 3)

Simplify

\frac{1}{2} [\frac{sec ^{2} ( u ) sin ( u )}{2}]_{0}^{a r c t a n (2 \cdot 2)} + \frac{1}{2} ln (22 + 3) : \frac{1}{2} ([\frac{1}{2} sec (u) tan (u)]_{0}^{a r c t a n (22)} + \frac{1}{2} ln (22 + 3))

= \frac{1}{2} ([\frac{1}{2} sec (u) tan (u)]_{0}^{a r c t a n (22)} + \frac{1}{2} ln (22 + 3))

Compute the boundaries:

32

= \frac{1}{2} (32 + \frac{1}{2} ln (22 + 3))

Simplify

= 3 + \frac{1}{2 2} ln (22 + 3)

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