What is the integral from 1 to 5 of sqrt(1+x^2) ?

The integral from 1 to 5 of sqrt(1+x^2) is (sqrt(2)(-1+5sqrt(13))-ln(1+sqrt(2))+ln(5+sqrt(26)))/2

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integral from 1 to 5 of sqrt(1+x^2)

\int_{1}^{5} 1 + x^{2} d x

\frac{2 ( - 1 + 5 13 ) - ln ( 1 + 2 ) + ln ( 5 + 26 )}{2}

Decimal

12.75597 \dots

Solution steps

\int_{1}^{5} 1 + x^{2} d x

Apply Trigonometric Substitution

: \int_{\frac{π}{4}}^{a r c t a n (5)} sec^{3} (u) d u

= \int_{\frac{π}{4}}^{a r c t a n (5)} sec^{3} (u) d u

Apply Integral Reduction

= [\frac{sec ^{2} ( u ) sin ( u )}{2}]_{\frac{π}{4}}^{a r c t a n (5)} + \frac{1}{2} \cdot \int_{\frac{π}{4}}^{a r c t a n (5)} sec (u) d u

\int_{\frac{π}{4}}^{a r c t a n (5)} sec (u) d u = ln (5 + 26) - ln (1 + 2)

= [\frac{sec ^{2} ( u ) sin ( u )}{2}]_{\frac{π}{4}}^{a r c t a n (5)} + \frac{1}{2} (ln (5 + 26) - ln (1 + 2))

Simplify

[\frac{sec ^{2} ( u ) sin ( u )}{2}]_{\frac{π}{4}}^{a r c t a n (5)} + \frac{1}{2} (ln (5 + 26) - ln (1 + 2)) : [\frac{1}{2} sec (u) tan (u)]_{\frac{π}{4}}^{a r c t a n (5)} + \frac{1}{2} (ln (5 + 26) - ln (1 + 2))

= [\frac{1}{2} sec (u) tan (u)]_{\frac{π}{4}}^{a r c t a n (5)} + \frac{1}{2} (ln (5 + 26) - ln (1 + 2))

Compute the boundaries:

\frac{5 13 - 1}{2}

= \frac{5 13 - 1}{2} + \frac{1}{2} (ln (5 + 26) - ln (1 + 2))

Simplify

= \frac{2 ( - 1 + 5 13 ) - ln ( 1 + 2 ) + ln ( 5 + 26 )}{2}

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What is the integral from 1 to 5 of sqrt(1+x^2) ?
The integral from 1 to 5 of sqrt(1+x^2) is (sqrt(2)(-1+5sqrt(13))-ln(1+sqrt(2))+ln(5+sqrt(26)))/2