Question

请问数学大神第五题234小题题怎么做

Answer 1

(2)令√x+1=t,x在0-3变成t在1-2
x+1=t^2
dx=2tdt
然后就是(t^2-1)2t/t+1,所以变成（t-1）2t=2(t^2-2t),所以变成2t^3/3-t^2在1到2的值
就是5/3
(3)和（4）都要化∫1/(1+x^2)dx这个形式的不定式，为arctanx+C
(3)令√x-1=0,t=0-1
dx=2tdt
2∫t^2/(t^2+1)dt
=2∫1-2∫1/(1+t^2)dt

2t-2arctant
2-π/2
(4)令t=√（e^x-1）,x=ln(t^2+1),dx=2t/(t^2+1),x=0-ln2,则t=0-1,
2∫t^2/(t^2+1)dt
=2∫1-2∫1/(1+t^2)dt
2t-2arctant
2-π/2

Answer 2

(2)=∫(1.2)(u²-1)/(1+u)d(u²-1)
=∫2u(u-1)du
=2u³/3-u²
=16/3-4-(2/3-1)
=5/3

追答

(3)=∫(0.1)u/(u²+1)d(u²+1)
=2∫1-1/(u²+1)du
=2u-2arctanu
=2-π/2

(4)=∫(0.1)udln(u²+1)
=∫2u²/(u²+1)du
=2u-2arctanu
=2-π/2