What is the integral from 1 to 3 of sqrt(x^2+1) ?

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

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integral from 1 to 3 of sqrt(x^2+1)

\int_{1}^{3} x^{2} + 1 d x

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

Decimal Notation

4.50484 \dots

Solution steps

\int_{1}^{3} x^{2} + 1 d x

Apply Trigonometric Substitution

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

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

Apply Integral Reduction

= [\frac{sec ^{2} ( u ) sin ( u )}{2}]_{\frac{π}{4}}^{a r c t a n (3)} + \frac{1}{2} \cdot \int_{\frac{π}{4}}^{a r c t a n (3)} sec (u) d u

\int_{\frac{π}{4}}^{a r c t a n (3)} sec (u) d u = ln (3 + 10) - ln (1 + 2)

= [\frac{sec ^{2} ( u ) sin ( u )}{2}]_{\frac{π}{4}}^{a r c t a n (3)} + \frac{1}{2} (ln (3 + 10) - ln (1 + 2))

Simplify

[\frac{sec ^{2} ( u ) sin ( u )}{2}]_{\frac{π}{4}}^{a r c t a n (3)} + \frac{1}{2} (ln (3 + 10) - ln (1 + 2)) : [\frac{1}{2} sec (u) tan (u)]_{\frac{π}{4}}^{a r c t a n (3)} + \frac{1}{2} (ln (3 + 10) - ln (1 + 2))

= [\frac{1}{2} sec (u) tan (u)]_{\frac{π}{4}}^{a r c t a n (3)} + \frac{1}{2} (ln (3 + 10) - ln (1 + 2))

Compute the boundaries:

\frac{3 5 - 1}{2}

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

Simplify

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

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