integral from 4 to x^2 of 3sqrt(1+t^2)

解答

\int_{4}^{x^{2}} 3 1 + t^{2} d t

3 (- 217 + \frac{x ^{2} 1 + x ^{4} + ln x ^{2} + 1 + x ^{4} - ln ( 4 + 17 )}{2})

求解步骤

\int_{4}^{x^{2}} 3 1 + t^{2} d t

提出常数:

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

= 3 \cdot \int_{4}^{x^{2}} 1 + t^{2} d t

使用三角换元法

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

使用降次积分法

= 3 [\frac{sec ^{2} ( u ) sin ( u )}{2}]_{a r c t a n (4)}^{a r c t a n (x^{2})} + \frac{1}{2} \cdot \int_{a r c t a n (4)}^{a r c t a n (x^{2})} sec (u) d u

\int_{a r c t a n (4)}^{a r c t a n (x^{2})} sec (u) d u = ln x^{2} + 1 + x^{4} - ln (4 + 17)

= 3 [\frac{sec ^{2} ( u ) sin ( u )}{2}]_{a r c t a n (4)}^{a r c t a n (x^{2})} + \frac{1}{2} (ln x^{2} + 1 + x^{4} - ln (4 + 17))

化简

3 [\frac{sec ^{2} ( u ) sin ( u )}{2}]_{a r c t a n (4)}^{a r c t a n (x^{2})} + \frac{1}{2} (ln x^{2} + 1 + x^{4} - ln (4 + 17)) : 3 ([\frac{1}{2} sec (u) tan (u)]_{a r c t a n (4)}^{a r c t a n (x^{2})} + \frac{1}{2} (ln x^{2} + 1 + x^{4} - ln (4 + 17)))

= 3 ([\frac{1}{2} sec (u) tan (u)]_{a r c t a n (4)}^{a r c t a n (x^{2})} + \frac{1}{2} (ln x^{2} + 1 + x^{4} - ln (4 + 17)))

带入上下限计算后得:

\frac{1}{2} x^{2} 1 + x^{4} - 217

= 3 (\frac{1}{2} x^{2} 1 + x^{4} - 217 + \frac{1}{2} (ln x^{2} + 1 + x^{4} - ln (4 + 17)))

化简

= 3 (- 217 + \frac{x ^{2} 1 + x ^{4} + ln x ^{2} + 1 + x ^{4} - ln ( 4 + 17 )}{2})

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