Question

(1+x^2)^(-3/2)的原函数是什么啊?

Answer 1

令x=tany,dx=sec²y dy

(1+x²)^(-3/2)=1/(1+tan²y)^(3/2)=1/(sec²y)^(3/2)=1/sec³y=cos³y

∴原式=∫cos³y*sec²y dy

=∫cos³y*1/cos²y dy

=∫cosy dy

=siny + C

=x/√(1+x²) + C,C为任意常数