What is the integral from 0 to 4 of sqrt(1+64x^2) ?

The integral from 0 to 4 of sqrt(1+64x^2) is 1/8 (80sqrt(41)+1/2 ln(32+5sqrt(41)))

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integral from 0 to 4 of sqrt(1+64x^2)

\int_{0}^{4} 1 + 64 x^{2} d x

\frac{1}{8} (8041 + \frac{1}{2} ln (32 + 541))

Decimal Notation

64.29118 \dots

Solution steps

\int_{0}^{4} 1 + 64 x^{2} d x

Apply Trigonometric Substitution

: \int_{0}^{a r c t a n (32)} \frac{1}{8} sec^{3} (u) d u

= \int_{0}^{a r c t a n (32)} \frac{1}{8} 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}{8} \cdot \int_{0}^{a r c t a n (32)} sec^{3} (u) d u

Apply Integral Reduction

= \frac{1}{8} ([\frac{sec ^{2} ( u ) sin ( u )}{2}]_{0}^{a r c t a n (32)} + \frac{1}{2} \cdot \int_{0}^{a r c t a n (32)} sec (u) d u)

\int_{0}^{a r c t a n (32)} sec (u) d u = ln (32 + 541)

= \frac{1}{8} ([\frac{sec ^{2} ( u ) sin ( u )}{2}]_{0}^{a r c t a n (32)} + \frac{1}{2} ln (32 + 541))

Simplify

\frac{1}{8} ([\frac{sec ^{2} ( u ) sin ( u )}{2}]_{0}^{a r c t a n (32)} + \frac{1}{2} ln (32 + 541)) : \frac{1}{8} ([\frac{1}{2} sec (u) tan (u)]_{0}^{a r c t a n (32)} + \frac{1}{2} ln (32 + 541))

= \frac{1}{8} ([\frac{1}{2} sec (u) tan (u)]_{0}^{a r c t a n (32)} + \frac{1}{2} ln (32 + 541))

Compute the boundaries:

8041

= \frac{1}{8} (8041 + \frac{1}{2} ln (32 + 541))

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What is the integral from 0 to 4 of sqrt(1+64x^2) ?
The integral from 0 to 4 of sqrt(1+64x^2) is 1/8 (80sqrt(41)+1/2 ln(32+5sqrt(41)))