What is the integral from 0 to 3 of sqrt(1+36x^2) ?

The integral from 0 to 3 of sqrt(1+36x^2) is 1/6 (45sqrt(13)+1/2 ln(18+5sqrt(13)))

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integral from 0 to 3 of sqrt(1+36x^2)

\int_{0}^{3} 1 + 36 x^{2} d x

\frac{1}{6} (4513 + \frac{1}{2} ln (18 + 513))

Decimal Notation

27.34032 \dots

Solution steps

\int_{0}^{3} 1 + 36 x^{2} d x

Apply Trigonometric Substitution

: \int_{0}^{a r c t a n (18)} \frac{1}{6} sec^{3} (u) d u

= \int_{0}^{a r c t a n (18)} \frac{1}{6} 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}{6} \cdot \int_{0}^{a r c t a n (18)} sec^{3} (u) d u

Apply Integral Reduction

= \frac{1}{6} ([\frac{sec ^{2} ( u ) sin ( u )}{2}]_{0}^{a r c t a n (18)} + \frac{1}{2} \cdot \int_{0}^{a r c t a n (18)} sec (u) d u)

\int_{0}^{a r c t a n (18)} sec (u) d u = ln (18 + 513)

= \frac{1}{6} ([\frac{sec ^{2} ( u ) sin ( u )}{2}]_{0}^{a r c t a n (18)} + \frac{1}{2} ln (18 + 513))

Simplify

\frac{1}{6} ([\frac{sec ^{2} ( u ) sin ( u )}{2}]_{0}^{a r c t a n (18)} + \frac{1}{2} ln (18 + 513)) : \frac{1}{6} ([\frac{1}{2} sec (u) tan (u)]_{0}^{a r c t a n (18)} + \frac{1}{2} ln (18 + 513))

= \frac{1}{6} ([\frac{1}{2} sec (u) tan (u)]_{0}^{a r c t a n (18)} + \frac{1}{2} ln (18 + 513))

Compute the boundaries:

4513

= \frac{1}{6} (4513 + \frac{1}{2} ln (18 + 513))

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What is the integral from 0 to 3 of sqrt(1+36x^2) ?
The integral from 0 to 3 of sqrt(1+36x^2) is 1/6 (45sqrt(13)+1/2 ln(18+5sqrt(13)))