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Find the indefinite integral. $$ \int \frac{x^{3}-3 x^{2}+4 x-9}{x^{2}+3} d x $$

Short Answer

Expert verified
\(\frac{1}{2}x^2 -3x - \ln|x^2 + 3| + C\) is the indefinite integral of the given function.

Step by step solution

01

Polynomial Division

First, conduct the polynomial division of \(x^{3}-3 x^{2}+4 x-9\) by \(x^{2}+3\). This simplifies the given expression: \(x - 3 + \frac{-2x+6}{x^2 + 3}\).
02

Integrating First Part

The first part of the simplified expression which is \(x - 3\), can be integrated directly using the power rule: \(\int x \, dx = \frac{1}{2}x^2\) and \(\int -3 \, dx = -3x\). Combining the two integrals, the result is \(\frac{1}{2}x^2 -3x + C1\), where \(C1\) is the constant of integration.
03

Substitution for Second Part

Next, let's move on to the second part of the expression, \(-2\int\frac{x-3}{x^2+3}dx\). This part can be solved by using the substitution method. Let \(u = x^2 + 3\), then \(du = 2x dx\). The integral becomes \(- \int \frac{du}{u}\). Integrate this to get \(- \ln|u| + C2\), where \(C2\) is the constant of integration.
04

Re-substitute and Simplify

Replace \(u\) back with the original expression \(x^2 + 3\) in the integral obtained in the previous step. Therefore the integral equates to \(- \ln|x^2 + 3| + C2\). So the integral of the entire function given is the sum of the results from step 2 and step 3 which is, \(\frac{1}{2}x^2 -3x - \ln|x^2 + 3| + C\), where \(C = C1+C2\).

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