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

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

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

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

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

Decimal

11.30527 \dots

Solution steps

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

Apply Trigonometric Substitution

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

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

Apply Integral Reduction

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

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

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

Simplify

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

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

Compute the boundaries:

310

= 310 + \frac{1}{2} (ln (3 + 10) - ln (- 3 + 10))

\frac{d y}{d x}, y - 9 x^{3} + 8 = 0, (2, 8)

4^{x^{2}}

\int (csc (2 x)) d x

y^{^{'}} - 2 t y = 12 t^{2} e^{t^{2}}, y (0) = 5

x \to 4 lim (\frac{x ^{2} - 4}{x - 2})

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