What is the integral from 0 to 4 of sqrt(1+4x^2) ?

The integral from 0 to 4 of sqrt(1+4x^2) is 1/2 (4sqrt(65)+1/2 ln(8+sqrt(65)))

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integral from 0 to 4 of sqrt(1+4x^2)

\int_{0}^{4} 1 + 4 x^{2} d x

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

Decimal Notation

16.81863 \dots

Solution steps

\int_{0}^{4} 1 + 4 x^{2} d x

Apply Trigonometric Substitution

: \int_{0}^{a r c t a n (8)} \frac{1}{2} sec^{3} (u) d u

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

Simplify

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

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

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What is the integral from 0 to 4 of sqrt(1+4x^2) ?
The integral from 0 to 4 of sqrt(1+4x^2) is 1/2 (4sqrt(65)+1/2 ln(8+sqrt(65)))