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

The integral from 2 to 8 of sqrt(x^2+1) is 4sqrt(65)-sqrt(5)+1/2 (ln(8+sqrt(65))-ln(2+sqrt(5)))

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

\int_{2}^{8} x^{2} + 1 d x

465 - 5 + \frac{1}{2} (ln (8 + 65) - ln (2 + 5))

Decimal Notation

30.67938 \dots

Solution steps

\int_{2}^{8} x^{2} + 1 d x

Apply Trigonometric Substitution

: \int_{a r c t a n (2)}^{a r c t a n (8)} sec^{3} (u) d u

= \int_{a r c t a n (2)}^{a r c t a n (8)} sec^{3} (u) d u

Apply Integral Reduction

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

\int_{a r c t a n (2)}^{a r c t a n (8)} sec (u) d u = ln (8 + 65) - ln (2 + 5)

= [\frac{sec ^{2} ( u ) sin ( u )}{2}]_{a r c t a n (2)}^{a r c t a n (8)} + \frac{1}{2} (ln (8 + 65) - ln (2 + 5))

Simplify

[\frac{sec ^{2} ( u ) sin ( u )}{2}]_{a r c t a n (2)}^{a r c t a n (8)} + \frac{1}{2} (ln (8 + 65) - ln (2 + 5)) : [\frac{1}{2} sec (u) tan (u)]_{a r c t a n (2)}^{a r c t a n (8)} + \frac{1}{2} (ln (8 + 65) - ln (2 + 5))

= [\frac{1}{2} sec (u) tan (u)]_{a r c t a n (2)}^{a r c t a n (8)} + \frac{1}{2} (ln (8 + 65) - ln (2 + 5))

Compute the boundaries:

465 - 5

= 465 - 5 + \frac{1}{2} (ln (8 + 65) - ln (2 + 5))

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What is the integral from 2 to 8 of sqrt(x^2+1) ?
The integral from 2 to 8 of sqrt(x^2+1) is 4sqrt(65)-sqrt(5)+1/2 (ln(8+sqrt(65))-ln(2+sqrt(5)))