Respuesta :
Answer:
[tex]\frac{2}{3}(\frac{1}{5}ln|3x-5|-\frac{1}{5}ln|x|)+C[/tex]
Step-by-step explanation:
We have been given a indefinite integral [tex]\int \frac{2}{3x\left(3x-5\right)}dx[/tex]. We are asked to find the indefinite integral.
We will use partial fraction formula to solve our given problem.
[tex]\frac{2}{3x\left(3x-5\right)}=\frac{3}{5(3x-5)}-\frac{1}{5x}[/tex]
[tex]\int \frac{2}{3x\left(3x-5\right)}dx=\frac{2}{3}\int \frac{1}{x\left(3x-5\right)}dx[/tex]
[tex]\frac{2}{3}\int \frac{1}{x\left(3x-5\right)}dx=\frac{2}{3}\int \frac{3}{5(3x-5)}-\frac{1}{5x}dx[/tex]
Using difference rule of integrals, we will get:
[tex]\frac{2}{3}(\int \frac{3}{5(3x-5)}dx-\int \frac{1}{5x}dx)[/tex]
Now, we need to use u-substitution as:
Let [tex]u=3x-5[/tex].
[tex]\frac{du}{dx}=3[/tex]
[tex]dx=\frac{1}{3}du[/tex]
[tex]\int \frac{3}{5(3x-5)}dx= \frac{3}{5}\int \frac{1}{(u)}*\frac{1}{3}du=\frac{3}{5}*\frac{1}{3}\int \frac{1}{(u)}du=\frac{1}{5}ln|u|=\frac{1}{5}ln|3x-5|[/tex]
[tex]\int \frac{1}{5x}dx=\frac{1}{5}\int \frac{1}{x}dx=\frac{1}{5}ln|x|[/tex]
Substitute back these values:
[tex]\frac{2}{3}(\int \frac{3}{5(3x-5)}dx-\int \frac{1}{5x}dx)=\frac{2}{3}(\frac{1}{5}ln|3x-5|-\frac{1}{5}ln|x|)[/tex]
Let us add a constant C.
[tex]\frac{2}{3}(\frac{1}{5}ln|3x-5|-\frac{1}{5}ln|x|)+C[/tex]
Therefore, our required integral would be [tex]\frac{2}{3}(\frac{1}{5}ln|3x-5|-\frac{1}{5}ln|x|)+C[/tex].
