What is the integral from 0 to 1 of sqrt(1+9x^2) ?

The integral from 0 to 1 of sqrt(1+9x^2) is sqrt(5/2)+1/6 ln(3+sqrt(10))

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integral from 0 to 1 of sqrt(1+9x^2)

\int_{0}^{1} 1 + 9 x^{2} d x

\frac{5}{2} + \frac{1}{6} ln (3 + 10)

Decimal Notation

1.88421 \dots

Solution steps

\int_{0}^{1} 1 + 9 x^{2} d x

Apply Trigonometric Substitution

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

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

Apply Integral Reduction

= \frac{1}{3} ([\frac{sec ^{2} ( u ) sin ( u )}{2}]_{0}^{a r c t a n (3)} + \frac{1}{2} \cdot \int_{0}^{a r c t a n (3)} sec (u) d u)

\int_{0}^{a r c t a n (3)} sec (u) d u = ln (3 + 10)

= \frac{1}{3} ([\frac{sec ^{2} ( u ) sin ( u )}{2}]_{0}^{a r c t a n (3)} + \frac{1}{2} ln (3 + 10))

Simplify

\frac{1}{3} ([\frac{sec ^{2} ( u ) sin ( u )}{2}]_{0}^{a r c t a n (3)} + \frac{1}{2} ln (3 + 10)) : \frac{1}{3} ([\frac{1}{2} sec (u) tan (u)]_{0}^{a r c t a n (3)} + \frac{1}{2} ln (3 + 10))

= \frac{1}{3} ([\frac{1}{2} sec (u) tan (u)]_{0}^{a r c t a n (3)} + \frac{1}{2} ln (3 + 10))

Compute the boundaries:

\frac{3 5}{2}

= \frac{1}{3} (\frac{3 5}{2} + \frac{1}{2} ln (3 + 10))

Simplify

= \frac{5}{2} + \frac{1}{6} ln (3 + 10)

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What is the integral from 0 to 1 of sqrt(1+9x^2) ?
The integral from 0 to 1 of sqrt(1+9x^2) is sqrt(5/2)+1/6 ln(3+sqrt(10))