integral from x to 5 of sqrt(1+t^2)

解答

\int_{x}^{5} 1 + t^{2} d t

\frac{5 13}{2} - \frac{x 1 + x ^{2}}{2} + \frac{1}{2} (ln (5 + 26) - ln x + 1 + x^{2})

求解步骤

\int_{x}^{5} 1 + t^{2} d t

使用三角换元法

= \int_{a r c t a n (x)}^{a r c t a n (5)} 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)}^{a r c t a n (5)} + \frac{1}{2} \cdot \int_{a r c t a n (x)}^{a r c t a n (5)} sec (u) d u

\int_{a r c t a n (x)}^{a r c t a n (5)} sec (u) d u = ln (5 + 26) - 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 (5)} + \frac{1}{2} (ln (5 + 26) - ln x + 1 + x^{2})

化简

[\frac{sec ^{2} ( u ) sin ( u )}{2}]_{a r c t a n (x)}^{a r c t a n (5)} + \frac{1}{2} (ln (5 + 26) - 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 (5)} + \frac{1}{2} (ln (5 + 26) - 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 (5)} + \frac{1}{2} (ln (5 + 26) - ln x + 1 + x^{2})

带入上下限计算后得:

\frac{5 13}{2} - \frac{1}{2} x 1 + x^{2}

= \frac{5 13}{2} - \frac{1}{2} x 1 + x^{2} + \frac{1}{2} (ln (5 + 26) - ln x + 1 + x^{2})

化简

= \frac{5 13}{2} - \frac{x 1 + x ^{2}}{2} + \frac{1}{2} (ln (5 + 26) - ln x + 1 + x^{2})

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