What is the integral from 0 to 1 of sqrt(1+8t^2) ?

The integral from 0 to 1 of sqrt(1+8t^2) is 3/2+1/(4sqrt(2))ln(2sqrt(2)+3)

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integral from 0 to 1 of sqrt(1+8t^2)

\int_{0}^{1} 1 + 8 t^{2} d t

\frac{3}{2} + \frac{1}{4 2} ln (22 + 3)

Decimal Notation

1.81161 \dots

Solution steps

\int_{0}^{1} 1 + 8 t^{2} d t

Apply Trigonometric Substitution

: \int_{0}^{a r c t a n (22)} \frac{1}{2 2} sec^{3} (u) d u

= \int_{0}^{a r c t a n (22)} \frac{1}{2 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 2} \cdot \int_{0}^{a r c t a n (22)} sec^{3} (u) d u

Apply Integral Reduction

= \frac{1}{2 2} [\frac{sec ^{2} ( u ) sin ( u )}{2}]_{0}^{a r c t a n (22)} + \frac{1}{2} \cdot \int_{0}^{a r c t a n (22)} sec (u) d u

\int_{0}^{a r c t a n (22)} sec (u) d u = ln (22 + 3)

= \frac{1}{2 2} [\frac{sec ^{2} ( u ) sin ( u )}{2}]_{0}^{a r c t a n (22)} + \frac{1}{2} ln (22 + 3)

Simplify

\frac{1}{2 2} [\frac{sec ^{2} ( u ) sin ( u )}{2}]_{0}^{a r c t a n (22)} + \frac{1}{2} ln (22 + 3) : \frac{1}{2 2} ([\frac{1}{2} sec (u) tan (u)]_{0}^{a r c t a n (22)} + \frac{1}{2} ln (22 + 3))

= \frac{1}{2 2} ([\frac{1}{2} sec (u) tan (u)]_{0}^{a r c t a n (22)} + \frac{1}{2} ln (22 + 3))

Compute the boundaries:

32

= \frac{1}{2 2} (32 + \frac{1}{2} ln (22 + 3))

Simplify

= \frac{3}{2} + \frac{1}{4 2} ln (22 + 3)

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What is the integral from 0 to 1 of sqrt(1+8t^2) ?
The integral from 0 to 1 of sqrt(1+8t^2) is 3/2+1/(4sqrt(2))ln(2sqrt(2)+3)