Answer:
[tex]\large\boxed{\int\left(1+\dfrac{4}{3x-1}+\dfrac{3}{x+2}\right)\ dx=x+\dfrac{4}{3}\ln(3x-1)+3\ln(x+2)}[/tex]
Step-by-step explanation:
[tex]\large{\int}\normal\left(1+\dfrac{4}{3x-1}+\dfrac{3}{x+2}\right)\ dx=\int1\ dx+\int\dfrac{4}{3x-1}\ dx+\int\dfrac{3}{x+2}\ dx\\\\(1)\int1\ dx=x\\\\(2)\int\dfrac{4}{3x-1}\ dx\Rightarrow\left|\begin{array}{ccc}3x-1=t\\3dx=dt\\dx=\frac{1}{3}dt\end{array}\right|\Rightarrow\int\dfrac{4}{3t}\ dt=\dfrac{4}{3}\int\dfrac{1}{t}\ dt=\dfrac{4}{3}\ln(t)=\dfrac{4}{3}\ln(3x-1)\\\\(3)\int\dfrac{3}{x+2}\ dx\Rightarrow\left|\begin{array}{ccc}x+2=u\\dx=du\end{array}\right|\Rightarrow\int\dfrac{3}{t}\ dt=3\int\dfrac{1}{t}\ dt=3\ln(t)=3\ln(x+2)[/tex]
[tex]\Downarrow\\\\\int\left(1+\dfrac{4}{3x-1}+\dfrac{3}{x+2}\right)\ dx=x+\dfrac{4}{3}\ln(3x-1)+3\ln(x+2)[/tex]
