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

解答

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

x 1 + 4 x^{2} + \frac{- 2 - ln ( 1 + 2 ) + ln 1 + 4 x ^{2} + 2 x}{2}

求解步骤

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

使用三角换元法

= \int_{\frac{π}{4}}^{a r c t a n (2 x)} sec^{3} (u) d u

使用降次积分法

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

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

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

化简

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

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

带入上下限计算后得:

x 1 + 4 x^{2} - \frac{1}{2}

= x 1 + 4 x^{2} - \frac{1}{2} + \frac{1}{2} (ln 2 x + 1 + 4 x^{2} - ln (1 + 2))

化简

= x 1 + 4 x^{2} + \frac{- 2 - ln ( 1 + 2 ) + ln 1 + 4 x ^{2} + 2 x}{2}

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