Question

高数,微积分,不定积分部分,分高,好好回答,

Answer 1

令x=a*tant, 则

1/(x√(x^2+a^2)) dx

=a*(sect)^2 / (a*tant*a*sect)

=1/a * csct.

因为csct原函数是-ln|csc(t)+cot(t)|+C.

所以原式的原函数是1/a *(-ln|csc(t)+cot(t)|)+C.

=1/a *(-ln|(√(x^2+a^2)+a)/x|)+C.

=1/a *(ln|x/(√(x^2+a^2)+a)|)+C.

=1/a *(ln|(√(x^2+a^2)-a)/x|)+C.