Question

Use the table of integrals, or a computer or calculator with symbolic integration capabilities, to find the indefinite integral.

Answer 1

Answer:

$\frac{2}{3}(\frac{1}{5}ln|3x-5|-\frac{1}{5}ln|x|)+C$

Step-by-step explanation:

We have been given a indefinite integral $\int \frac{2}{3x\left(3x-5\right)}dx$. We are asked to find the indefinite integral.

We will use partial fraction formula to solve our given problem.

$\frac{2}{3x\left(3x-5\right)}=\frac{3}{5(3x-5)}-\frac{1}{5x}$

$\int \frac{2}{3x\left(3x-5\right)}dx=\frac{2}{3}\int \frac{1}{x\left(3x-5\right)}dx$

$\frac{2}{3}\int \frac{1}{x\left(3x-5\right)}dx=\frac{2}{3}\int \frac{3}{5(3x-5)}-\frac{1}{5x}dx$

Using difference rule of integrals, we will get:

$\frac{2}{3}(\int \frac{3}{5(3x-5)}dx-\int \frac{1}{5x}dx)$

Now, we need to use u-substitution as:

Let $u=3x-5$.

$\frac{du}{dx}=3$

$dx=\frac{1}{3}du$

$\int \frac{3}{5(3x-5)}dx= \frac{3}{5}\int \frac{1}{(u)}*\frac{1}{3}du=\frac{3}{5}*\frac{1}{3}\int \frac{1}{(u)}du=\frac{1}{5}ln|u|=\frac{1}{5}ln|3x-5|$

$\int \frac{1}{5x}dx=\frac{1}{5}\int \frac{1}{x}dx=\frac{1}{5}ln|x|$

Substitute back these values:

$\frac{2}{3}(\int \frac{3}{5(3x-5)}dx-\int \frac{1}{5x}dx)=\frac{2}{3}(\frac{1}{5}ln|3x-5|-\frac{1}{5}ln|x|)$

Let us add a constant C.

$\frac{2}{3}(\frac{1}{5}ln|3x-5|-\frac{1}{5}ln|x|)+C$

Therefore, our required integral would be $\frac{2}{3}(\frac{1}{5}ln|3x-5|-\frac{1}{5}ln|x|)+C$.