Question

∫[(3x+1)/(4+x^2)^(1/2)]dx ∫1/[(x^2)(4-x^2)]dx

Answer 1

（1）

原式=∫[3x/(4+x^2)^(1/2)]dx+∫dx/(4+x^2)^(1/2)

=3/2*∫d(4+x^2)/(4+x^2)^(1/2)+ln[x+(4+x^2)^(1/2)]+C

=3(4+x^2)^(1/2)+ln[x+(4+x^2)^(1/2)]+C

(2)

原式=1/4*∫[1/x^2+1/(4-x^2)]dx

=1/4*[∫dx/x^2+∫dx/(4-x^2)]

=1/4*[-1/x-1/4*ln|(x-2)/(x+2)|]+C

=[ln|(x+2)/(x-2)|]/16-1/4x+C