Question

Use partial fractions to find the indefinite integral. $$ \int \frac{3 x}{x^{2}-6 x+9} d x $$

Short answer

The indefinite integral of the given function is \(3 \ln |x - 3| + \frac{9}{x - 3} + C\).

Step-by-step solution

Step 1. Factorize the denominator

We can rewrite the denominator \(x^{2} - 6x + 9\) in the form of \((x - a)^{2}\), where \(a\) is a real number. So the denominator becomes \((x - 3)^{2}\). The integral now looks as follows: \[\int \frac{3x}{(x - 3)^{2}} dx \]

Step 2. Divide numerator by the denominator to obtain partial fractions

Because the numerator's degree is less than the denominator's, we will not perform long division. Immediately proceed to partial fraction decomposition. The integral now becomes:\[\int \frac{3}{x - 3} dx - \int \frac{9}{(x - 3)^{2}} dx\]

Step 3. Evaluate the integral

We can now integrate term by term using the rules of integral calculus. The first integral is easy to compute because it is of the form \(\int \frac{1}{x} dx\), which is \(ln|x|\). The second integral is the integral of a square root in the denominator, and it is also known. This gives the following calculation:\[3 \int \frac{1}{x - 3} dx - 9 \int \frac{1}{(x - 3)^{2}} dx = 3 \ln |x - 3| + \frac{9}{x - 3} + C\]The 'C' denotes the constant of integration. This is because indefinite integration represents a family of functions, and all of these functions will differentiate to give the original function.