What is the integral from 0 to 3 of sqrt(1+16x^2) ?

The integral from 0 to 3 of sqrt(1+16x^2) is 1/4 (6sqrt(145)+1/2 ln(12+sqrt(145)))

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integral from 0 to 3 of sqrt(1+16x^2)

\int_{0}^{3} 1 + 16 x^{2} d x

\frac{1}{4} (6145 + \frac{1}{2} ln (12 + 145))

Decimal

18.45986 \dots

Solution steps

\int_{0}^{3} 1 + 16 x^{2} d x

Apply Trigonometric Substitution

= \int_{0}^{a r c t a n (12)} \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 (12)} 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 (12)} + \frac{1}{2} \cdot \int_{0}^{a r c t a n (12)} sec (u) d u)

\int_{0}^{a r c t a n (12)} sec (u) d u = ln (12 + 145)

= \frac{1}{4} ([\frac{sec ^{2} ( u ) sin ( u )}{2}]_{0}^{a r c t a n (12)} + \frac{1}{2} ln (12 + 145))

Simplify

\frac{1}{4} ([\frac{sec ^{2} ( u ) sin ( u )}{2}]_{0}^{a r c t a n (12)} + \frac{1}{2} ln (12 + 145)) : \frac{1}{4} ([\frac{1}{2} sec (u) tan (u)]_{0}^{a r c t a n (12)} + \frac{1}{2} ln (12 + 145))

= \frac{1}{4} ([\frac{1}{2} sec (u) tan (u)]_{0}^{a r c t a n (12)} + \frac{1}{2} ln (12 + 145))

Compute the boundaries:

6145

= \frac{1}{4} (6145 + \frac{1}{2} ln (12 + 145))

What is the integral from 0 to 3 of sqrt(1+16x^2) ?
The integral from 0 to 3 of sqrt(1+16x^2) is 1/4 (6sqrt(145)+1/2 ln(12+sqrt(145)))