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

解

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

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

解答ステップ

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

三角関数による置換を適用する

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

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

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

簡素化

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

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

限度を計算する:

- \frac{1}{2} - \frac{1}{2} x 1 + x^{2}

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

簡素化

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