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

解

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

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

解答ステップ

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

三角関数による置換を適用する

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

積分換算を適用する

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

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

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

簡素化

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

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

限度を計算する:

x 1 + x^{2}

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