微积分题。高手请进。

Question

∫[√(x^2+b^2)_√(x^2+a^2)]dx=?注意，积分号是从0积到正无穷，是定积分，这里我打不出来，请大神来帮忙。问题补充：

b>a
答对者十分感谢。急急急急！！！！

Answer 1

∫sec³(x)dx

= ∫sec(x)sec²(x)dx

∫udv = uv - ∫vdu 

u = sec(x)

du = sec(x)tan(x)dx

dv = sec²(x)dx

v = tan(x)

∫sec³(x)dx

= ∫sec(x)sec²(x)dx

= sec(x)tan(x) - ∫tan(x)sec(x)tan(x)dx

= sec(x)tan(x) - ∫tan²(x)sec(x)dx

= sec(x)tan(x) - ∫((sec²(x) - 1)sec(x))dx

= sec(x)tan(x) - ∫(sec³(x) - sec(x))dx

= sec(x)tan(x) - ∫sec³(x)dx + ∫sec(x)dx

= sec(x)tan(x) + ln(sec(x) + tan(x)) + c1 - ∫sec³(x)dx

2∫sec³(x)dx = sec(x)tan(x) + ln(sec(x) + tan(x)) + c1

∫sec³(x)dx = (1/2)(sec(x)tan(x) + ln(sec(x) + tan(x))) + c

Answer 2

∫√(x^2+b^2)dx
=x√(x^2+b^2)-∫x^2dx/√(x^2+b^2)
=x√(x^2+b^2)+b^2∫d((x/b)/√((x/b)^2+1)-∫√(x^2+b^2)dx
2∫√(x^2+b^2)dx=x√(x^2+b^2)+b^2∫d(x/b)/√((x/b)^2+1)
∫√(x^2+b^2)dx=(x/2)√(x^2+b^2)+(b^2/2)ln|x^2+b^2|+C
x/b=tanu ∫d(x/b)/√((x/b)^2+1)=∫secudu=ln|secu+tanu|+C=ln|x+√(x^2+b^2)|

∫[0,+∝]√(x^2+b^2)-√(x^2+a^2) dx
=lim(m->+∝)(m/2)[√(m^2+b^2)-√(m^2+a^2)]+(b^2/2)ln|m^2+b^2|-(a^2/2)ln|m^2+a^2|
+(b^2/2)ln|b^2|-(a^2/2)ln|a^2|
=(1/4)(b^2-a^2)+(b^2/2)ln|b^2|-(a^2/2)ln|a^2|
lim(m->+∝)(m/2)[√(m^2+b^2)-√(m^2+a^2)]
=lim(m->+∝)(1/2)(b^2-a^2)/[√(1+(b/m)^2)+√(1+(a/m)^2)]
=(1/4)(b^2-a^2)
lim(m->+∝)(b^2/2)ln|m^2+b^2|-(a^2/2)ln|m^2+a^2|
=lim(m->+∝)[(b^2/2)ln|1+b^2/m|-(a^2/2)ln|1+a^2/m^2|+ (b^2ln|m|-a^2ln|m|)
=0 lim(x->+∝)b^2ln|m|=lim(x->+∝)a^2ln|m|

Answer 3

可惜我数值计算早忘了