Evaluate the integral: I = integral (x³)/(x+3) dx
Evaluate the integral: $$ I = \int \frac{x^3}{x+3}\,dx. $$
Explanation of Method
Since the numerator's degree exceeds the denominator's, polynomial long division reduces the integrand to a polynomial plus a simple remainder term, both of which integrate directly.
- Divide $x^3$ by $x+3$ using polynomial long division.
- Rewrite the integral as a polynomial plus a remainder fraction.
- Integrate term by term.
Step 1 · Polynomial long division
Dividing $x^3$ by $x+3$:
$$ \frac{x^3}{x+3} = x^2 - 3x + 9 - \frac{27}{x+3}. $$
This can be verified by multiplying back: $(x+3)(x^2-3x+9) = x^3+27$, so $x^3 = (x+3)(x^2-3x+9) - 27$.
Step 2 · Rewrite the integral
Substituting the division result:
$$ I = \int \left(x^2 - 3x + 9 - \frac{27}{x+3}\right) dx. $$
Step 3 · Integrate term by term
Integrating each term:
$$ \int x^2\,dx = \frac{x^3}{3}, \qquad \int -3x\,dx = -\frac{3x^2}{2}, \qquad \int 9\,dx = 9x, \qquad \int \frac{27}{x+3}\,dx = 27\ln|x+3|. $$
Step 4 · Combine the results
Adding all the pieces together:
$$ I = \frac{x^3}{3} - \frac{3x^2}{2} + 9x - 27\ln|x+3| + C. $$
Summary Table
| Term | Antiderivative |
|---|---|
| $x^2$ | $x^3/3$ |
| $-3x$ | $-3x^2/2$ |
| $9$ | $9x$ |
| $-27/(x+3)$ | $-27\ln|x+3|$ |
Additional Notes
- The domain excludes $x=-3$, where the original integrand is undefined.
- Whenever a rational function's numerator has degree $\geq$ the denominator's degree, polynomial long division should be the first step — it isolates the "improper" part as a polynomial, leaving a proper fraction that is easy to integrate.
- The result can be checked by differentiating $\tfrac{x^3}{3}-\tfrac{3x^2}{2}+9x-27\ln|x+3|$ and confirming it returns $\dfrac{x^3}{x+3}$.
Result
The solution to the integral is: