Question

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

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:

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