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

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

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

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

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

Solution steps

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

Apply Trigonometric Substitution

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

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

Apply Integral Reduction

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

\int_{a r c t a n (- y)}^{a r c t a n (y)} sec (u) d u = ln y + 1 + y^{2} - ln - y + 1 + y^{2}

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

Simplify

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

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

Compute the boundaries:

y 1 + y^{2}

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

\int_{0}^{7} (49 - x^{2}) d x

\int_{0}^{2} 2 x^{3} - 6 x^{2} + 4 x d x

\int_{0}^{1} x (2 - x) d x

\int_{\frac{π}{4}}^{\frac{π}{2}} sec^{2} (x) d x

\int_{0}^{1} \frac{1}{2 x + 1} d x

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