integral from 5 to x^2 of 4sqrt(1+x^2)

解答

\int_{5}^{x^{2}} 4 1 + x^{2} d x

4 (\frac{x ^{2} 1 + x ^{4}}{2} - \frac{5 13}{2} + \frac{1}{2} (ln x^{2} + 1 + x^{4} - ln (5 + 26)))

求解步骤

\int_{5}^{x^{2}} 4 1 + x^{2} d x

提出常数:

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

= 4 \cdot \int_{5}^{x^{2}} 1 + x^{2} d x

使用三角换元法

= 4 \cdot \int_{a r c t a n (5)}^{a r c t a n (x^{2})} sec^{3} (u) d u

使用降次积分法

= 4 [\frac{sec ^{2} ( u ) sin ( u )}{2}]_{a r c t a n (5)}^{a r c t a n (x^{2})} + \frac{1}{2} \cdot \int_{a r c t a n (5)}^{a r c t a n (x^{2})} sec (u) d u

\int_{a r c t a n (5)}^{a r c t a n (x^{2})} sec (u) d u = ln x^{2} + 1 + x^{4} - ln (5 + 26)

= 4 [\frac{sec ^{2} ( u ) sin ( u )}{2}]_{a r c t a n (5)}^{a r c t a n (x^{2})} + \frac{1}{2} (ln x^{2} + 1 + x^{4} - ln (5 + 26))

化简

4 [\frac{sec ^{2} ( u ) sin ( u )}{2}]_{a r c t a n (5)}^{a r c t a n (x^{2})} + \frac{1}{2} (ln x^{2} + 1 + x^{4} - ln (5 + 26)) : 4 ([\frac{1}{2} sec (u) tan (u)]_{a r c t a n (5)}^{a r c t a n (x^{2})} + \frac{1}{2} (ln x^{2} + 1 + x^{4} - ln (5 + 26)))

= 4 ([\frac{1}{2} sec (u) tan (u)]_{a r c t a n (5)}^{a r c t a n (x^{2})} + \frac{1}{2} (ln x^{2} + 1 + x^{4} - ln (5 + 26)))

带入上下限计算后得:

\frac{1}{2} x^{2} 1 + x^{4} - \frac{5 13}{2}

= 4 (\frac{1}{2} x^{2} 1 + x^{4} - \frac{5 13}{2} + \frac{1}{2} (ln x^{2} + 1 + x^{4} - ln (5 + 26)))

化简

= 4 (\frac{x ^{2} 1 + x ^{4}}{2} - \frac{5 13}{2} + \frac{1}{2} (ln x^{2} + 1 + x^{4} - ln (5 + 26)))

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