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

解答

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

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

求解步骤

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

使用三角换元法

= \int_{a r c t a n (x)}^{- \frac{π}{4}} sec^{3} (u) d u

使用降次积分法

= [\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})

化简

[\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})

带入上下限计算后得:

- \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})

化简

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

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