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受欢迎的微积分 >

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

解答

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

解答

3 + ln (\frac{1}{2} + \frac{3}{2}) - ln (- \frac{1}{2} + \frac{3}{2})

+1

十进制

3.04900 \dots

求解步骤

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

使用三角换元法

= \int_{- a r c t a n (\frac{1}{2})}^{a r c t a n (\frac{1}{2})} 2 sec^{3} (u) d u

提出常数:

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

= 2 \cdot \int_{- a r c t a n (\frac{1}{2})}^{a r c t a n (\frac{1}{2})} sec^{3} (u) d u

使用降次积分法

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

\int_{- a r c t a n (\frac{1}{2})}^{a r c t a n (\frac{1}{2})} sec (u) d u = ln (\frac{1}{2} + \frac{3}{2}) - ln (- \frac{1}{2} + \frac{3}{2})

= 2 [\frac{sec ^{2} ( u ) sin ( u )}{2}]_{- a r c t a n (\frac{1}{2})}^{a r c t a n (\frac{1}{2})} + \frac{1}{2} (ln (\frac{1}{2} + \frac{3}{2}) - ln (- \frac{1}{2} + \frac{3}{2}))

化简

2 [\frac{sec ^{2} ( u ) sin ( u )}{2}]_{- a r c t a n (\frac{1}{2})}^{a r c t a n (\frac{1}{2})} + \frac{1}{2} (ln (\frac{1}{2} + \frac{3}{2}) - ln (- \frac{1}{2} + \frac{3}{2})) : 2 [\frac{1}{2} sec (u) tan (u)]_{- a r c t a n (\frac{1}{2})}^{a r c t a n (\frac{1}{2})} + ln (\frac{1}{2} + \frac{3}{2}) - ln (- \frac{1}{2} + \frac{3}{2})

= 2 [\frac{1}{2} sec (u) tan (u)]_{- a r c t a n (\frac{1}{2})}^{a r c t a n (\frac{1}{2})} + ln (\frac{1}{2} + \frac{3}{2}) - ln (- \frac{1}{2} + \frac{3}{2})

带入上下限计算后得:

\frac{3}{2}

= 2 \cdot \frac{3}{2} + ln (\frac{1}{2} + \frac{3}{2}) - ln (- \frac{1}{2} + \frac{3}{2})

化简

= 3 + ln (\frac{1}{2} + \frac{3}{2}) - ln (- \frac{1}{2} + \frac{3}{2})

作图

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