74. $\\int(1+\\frac{1}{x})\\cot(x+\\ln x)dx$

Explanation:

Step1: Let (u = x+\ln x)

Differentiate (u) with respect to (x). Using the sum - rule ((f + g)^\prime=f^\prime+g^\prime), where (f(x)=x) and (g(x)=\ln x). The derivative of (y = x) is (y^\prime=1), and the derivative of (y=\ln x) is (y^\prime=\frac{1}{x}). So, (du=(1 + \frac{1}{x})dx).

Step2: Substitute (u) and (du) into the integral

The integral (\int(1+\frac{1}{x})\cot(x + \ln x)dx) becomes (\int\cot udu). We know that (\int\cot udu=\int\frac{\cos u}{\sin u}du). Let (t=\sin u), then (dt=\cos udu). So (\int\frac{\cos u}{\sin u}du=\int\frac{dt}{t}).

Step3: Integrate (\int\frac{dt}{t})

Using the formula (\int\frac{1}{t}dt=\ln|t|+C). Since (t = \sin u), the integral (\int\frac{dt}{t}=\ln|\sin u|+C).

Step4: Substitute back (u=x+\ln x)

We get (\ln|\sin(x+\ln x)|+C).

Answer:

(\ln|\sin(x+\ln x)|+C)