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

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

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

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

2 + \frac{1}{2} (ln (1 + 2) - ln (- 1 + 2))

Decimal Notation

2.29558 \dots

Solution steps

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

Apply Trigonometric Substitution

: \int_{- \frac{π}{4}}^{\frac{π}{4}} sec^{3} (u) d u

= \int_{- \frac{π}{4}}^{\frac{π}{4}} sec^{3} (u) d u

Apply Integral Reduction

= [\frac{sec ^{2} ( u ) sin ( u )}{2}]_{- \frac{π}{4}}^{\frac{π}{4}} + \frac{1}{2} \cdot \int_{- \frac{π}{4}}^{\frac{π}{4}} sec (u) d u

\int_{- \frac{π}{4}}^{\frac{π}{4}} sec (u) d u = ln (1 + 2) - ln (- 1 + 2)

= [\frac{sec ^{2} ( u ) sin ( u )}{2}]_{- \frac{π}{4}}^{\frac{π}{4}} + \frac{1}{2} (ln (1 + 2) - ln (- 1 + 2))

Simplify

[\frac{sec ^{2} ( u ) sin ( u )}{2}]_{- \frac{π}{4}}^{\frac{π}{4}} + \frac{1}{2} (ln (1 + 2) - ln (- 1 + 2)) : [\frac{1}{2} sec (u) tan (u)]_{- \frac{π}{4}}^{\frac{π}{4}} + \frac{1}{2} (ln (1 + 2) - ln (- 1 + 2))

= [\frac{1}{2} sec (u) tan (u)]_{- \frac{π}{4}}^{\frac{π}{4}} + \frac{1}{2} (ln (1 + 2) - ln (- 1 + 2))

Compute the boundaries:

2

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

(sin (t)) y^{^{'}} + (cos (t)) y = 3 t (sin (t))

f (x) = \frac{4 x + 2}{4 x - 2}

arctan (t)

x \to - 1 lim ((x + 2)^{2})

sin^{2} (x) cos^{2} (x)

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