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

The integral from 0 to 1 of sqrt(16x^2+1) is 1/4 (2sqrt(17)+1/2 ln(4+sqrt(17)))

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

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

\frac{1}{4} (217 + \frac{1}{2} ln (4 + 17))

Solution steps

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

Apply Trigonometric Substitution

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

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

\int_{0}^{a r c t a n (4)} sec (u) d u = ln (4 + 17)

= \frac{1}{4} ([\frac{sec ^{2} ( u ) sin ( u )}{2}]_{0}^{a r c t a n (4)} + \frac{1}{2} ln (4 + 17))

Simplify

\frac{1}{4} ([\frac{sec ^{2} ( u ) sin ( u )}{2}]_{0}^{a r c t a n (4)} + \frac{1}{2} ln (4 + 17)) : \frac{1}{4} ([\frac{1}{2} sec (u) tan (u)]_{0}^{a r c t a n (4)} + \frac{1}{2} ln (4 + 17))

= \frac{1}{4} ([\frac{1}{2} sec (u) tan (u)]_{0}^{a r c t a n (4)} + \frac{1}{2} ln (4 + 17))

Compute the boundaries:

217

= \frac{1}{4} (217 + \frac{1}{2} ln (4 + 17))

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What is the integral from 0 to 1 of sqrt(16x^2+1) ?
The integral from 0 to 1 of sqrt(16x^2+1) is 1/4 (2sqrt(17)+1/2 ln(4+sqrt(17)))