integral from 0 to x of sqrt(t^4+t^6)

解答

\int_{0}^{x} t^{4} + t^{6} d t

- \frac{x x ^{2} + 1 + ln x + x ^{2} + 1}{2} + \frac{x ( x ^{2} + 1 ) ^{\frac{3}{2}}}{4} + \frac{3 ( x x ^{2} + 1 + ln x + x ^{2} + 1 )}{8}

求解步骤

\int_{0}^{x} t^{4} + t^{6} d t

化简

t^{4} + t^{6} : t^{2} 1 + t^{2}

= \int_{0}^{x} t^{2} 1 + t^{2} d t

使用三角换元法

= \int_{0}^{a r c t a n (x)} sec^{3} (u) tan^{2} (u) d u

使用三角恒等式改写

= \int_{0}^{a r c t a n (x)} sec^{3} (u) (- 1 + sec^{2} (u)) d u

乘开

sec^{3} (u) (- 1 + sec^{2} (u)) : - sec^{3} (u) + sec^{5} (u)

= \int_{0}^{a r c t a n (x)} - sec^{3} (u) + sec^{5} (u) d u

使用积分加法定则:

\int f (x) \pm g (x) d x = \int f (x) d x \pm \int g (x) d x

= - \int_{0}^{a r c t a n (x)} sec^{3} (u) d u + \int_{0}^{a r c t a n (x)} sec^{5} (u) d u

\int_{0}^{a r c t a n (x)} sec^{3} (u) d u = \frac{x x ^{2} + 1 + ln x + x ^{2} + 1}{2}

\int_{0}^{a r c t a n (x)} sec^{5} (u) d u = \frac{x ( x ^{2} + 1 ) ^{\frac{3}{2}}}{4} + \frac{3 ( x x ^{2} + 1 + ln x + x ^{2} + 1 )}{8}

= - \frac{x x ^{2} + 1 + ln x + x ^{2} + 1}{2} + \frac{x ( x ^{2} + 1 ) ^{\frac{3}{2}}}{4} + \frac{3 ( x x ^{2} + 1 + ln x + x ^{2} + 1 )}{8}

y^{^{''}} - 2 y^{^{'}} - 2 y = 0

x \to n - lim ([x])

\frac{d}{d x} ((3 - e^{x})^{8})

\frac{d}{d x} (180 - \frac{x ^{2}}{45})

d er i v a t i v e f (x) = 8 1 + x