Question

求不定积分 ∫(x∧2+a∧2)∧3/2 dx

Answer 1

令x=atanu,则dx=a(secu)^2 du
∫(x^2+a^2)^(3/2) dx
=∫ (a^4)(secu)^3·(secu)^2 du
=(a^4)∫ (secu)^5 du
=(a^4)[1/4·tanu(secu)^3+3/4·∫(secu)^3 du]
=(a^4)[1/4·tanu(secu)^3+3/8·tanu·secu+3/8·∫secu du]
=(a^4)[1/4·tanu(secu)^3+3/8·tanu·secu+3/8·ln|secu+tanu|]+C
=1/4·x·(a^2+x^2)^(3/2)+3/8·(a^2)·x·(a^2+x^2)^(1/2)+3/8·(a^4)·ln|(x^2/a^2+1)^(1/2)+x/a|]+C
注：∫(secx)^m dx=[sinx·(secx)^(m-1)]/(m-1)+(m-2)/(m-1)·∫(secx)^(m-2)dx