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

解答

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

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

求解步骤

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

提出常数:

\int a \cdot f (x) d x = a \cdot \int f (x) d x

= x^{2} \cdot \int_{x}^{1} 1 + u^{2} d u

使用三角换元法

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

使用降次积分法

= x^{2} ([\frac{sec ^{2} ( v ) sin ( v )}{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 (v) d v)

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

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

化简

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

= x^{2} ([\frac{1}{2} sec (v) tan (v)]_{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}

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

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