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

解答

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

x^{2} 1 + x^{4} - 526 + ln x^{2} + 1 + x^{4} - ln (5 + 26)

求解步骤

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

提出常数:

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

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

使用三角换元法

= 2 \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

= 2 [\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)

= 2 [\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

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

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

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

化简

= x^{2} 1 + x^{4} - 526 + ln x^{2} + 1 + x^{4} - ln (5 + 26)

\int_{π}^{2 π} (θ) d θ

\int_{9}^{\infty} \frac{1}{x} d x

\int_{0}^{6} e^{x} - e^{- x} d x

\int_{5}^{26} x e^{- x^{2}} d x

\int_{1}^{8} 1 + 4 x^{- \frac{2}{3}} d x