Помогите решить с помощью замены
∫[∛(ln(3x+1))/(3x+1)] dx= I t=ln(3x+1) dt=[1/(3x+1)]3dx I ⇔ ∫[∛(ln(3x+1))/(3x+1)] dx=(1/3)∫∛t dt =(1/3)·[ t^(1/3+1)] /(1/3+1)+C=[t^(4/3)]/4+C= {ln(3x+1)]^(4/3)}/4+C