Question

高数求不定积分，高悬赏，尽量详细点，速度

Answer 1

1. 令x^0.5 =t , 则x = t^2, dx = 2*t*dt, 代入积分得，2*(x^0.5 - ln(x^0.5 + 1)) + C. 2. 令(2-x)^0.5 = t, 则x = 2-t^2, dx = -2*t*dt, 代入积分得，-(2*(2 - x)^(1/2)*(20*x + 3*(x - 2)^2 + 20))/15 +C 3. 令 x = sec(t) , 则dx = sec(t)*tan(t)*dt，代入积分得arccos(1/x) + C. 4. 令x = a*sin(t) ,则dx = a*cos(t)*dt, 代入积分得，-arcsin(x/a) - cot(arcsin(x/a)) + C. 5. 令 x = sin(t),则dx = cos(t)*dt,代入得cos(t)/(1+cos(t))的定积分，再令 u = tan(t/2),则 cos(t) = (1-u^2)/(1+u^2), dt = 2/(1+u^2)*du,代入得 cos(t)/(1+cos(t))的定积分为t - tan(t/2) ,再代入x = sin(t) 得arcsin(x) - tan(arcsin(x)/2) + C。

Answer 2

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