What is the integral from 0 to 2 of sqrt(1+16x^2) ?

The integral from 0 to 2 of sqrt(1+16x^2) is sqrt(65)+1/8 ln(8+sqrt(65))

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integral from 0 to 2 of sqrt(1+16x^2)

\int_{0}^{2} 1 + 16 x^{2} d x

65 + \frac{1}{8} ln (8 + 65)

Decimal

8.40931 \dots

Solution steps

\int_{0}^{2} 1 + 16 x^{2} d x

Apply Trigonometric Substitution

: \int_{0}^{a r c t a n (8)} \frac{1}{4} sec^{3} (u) d u

= \int_{0}^{a r c t a n (8)} \frac{1}{4} 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}{4} \cdot \int_{0}^{a r c t a n (8)} sec^{3} (u) d u

Apply Integral Reduction

= \frac{1}{4} ([\frac{sec ^{2} ( u ) sin ( u )}{2}]_{0}^{a r c t a n (8)} + \frac{1}{2} \cdot \int_{0}^{a r c t a n (8)} sec (u) d u)

\int_{0}^{a r c t a n (8)} sec (u) d u = ln (8 + 65)

= \frac{1}{4} ([\frac{sec ^{2} ( u ) sin ( u )}{2}]_{0}^{a r c t a n (8)} + \frac{1}{2} ln (8 + 65))

Simplify

\frac{1}{4} ([\frac{sec ^{2} ( u ) sin ( u )}{2}]_{0}^{a r c t a n (8)} + \frac{1}{2} ln (8 + 65)) : \frac{1}{4} ([\frac{1}{2} sec (u) tan (u)]_{0}^{a r c t a n (8)} + \frac{1}{2} ln (8 + 65))

= \frac{1}{4} ([\frac{1}{2} sec (u) tan (u)]_{0}^{a r c t a n (8)} + \frac{1}{2} ln (8 + 65))

Compute the boundaries:

465

= \frac{1}{4} (465 + \frac{1}{2} ln (8 + 65))

Simplify

= 65 + \frac{1}{8} ln (8 + 65)

\int \frac{1}{x ^{15}} d x

x \to π lim (5 \frac{x - π}{x + π})

\frac{d y}{d x} + 4 y = 60

2 π \frac{x}{g}

\frac{d}{d x} (\frac{- ln ( x ) + 1}{x ^{2}})

What is the integral from 0 to 2 of sqrt(1+16x^2) ?
The integral from 0 to 2 of sqrt(1+16x^2) is sqrt(65)+1/8 ln(8+sqrt(65))