Evaluate the integral: I = integral ((x+3) dx)/(x² + 6x
Evaluate the integral: $$ I = \int \frac{(x+3)\,dx}{x^2 + 6x - 5}. $$
Explanation of Method
Unlike Problem 67, the numerator here is only proportional to (not exactly equal to) the derivative of the denominator: $\tfrac{d}{dx}(x^2+6x-5)=2x+6=2(x+3)$. Recognizing this factor of $2$ lets the same logarithmic substitution apply, with an extra $\tfrac12$ carried through.
- Factor the denominator (for insight).
- Compare the numerator to the denominator's derivative, noting the proportionality constant.
- Substitute and integrate using the logarithmic antiderivative, adjusted by that constant.
- Substitute back and add the constant of integration.
Step 1 · Factor the denominator and compare with the numerator
To solve this integral, we first factorize the denominator:
$$ x^2+6x-5 = (x+5)(x-1). $$
Next, we check if the numerator $x+3$ can be rewritten in terms of the derivative of the denominator. The derivative of the denominator is:
$$ \frac{d}{dx}\left(x^2+6x-5\right) = 2x+6 = 2(x+3). $$
We notice that $x+3$ is not directly the derivative but suggests a substitution method. We set:
$$ u = x^2+6x-5. $$
Differentiating both sides:
$$ \frac{du}{dx} = 2x+6. $$
Rearranging:
$$ du = (2x+6)\,dx = 2(x+3)\,dx. $$
Dividing by $2$:
$$ \frac{du}{2} = (x+3)\,dx. $$
Step 2 · Integrate
Now, substituting into the integral:
$$ I = \int \frac{\frac{du}{2}}{u}. $$
Factoring out the constant:
$$ I = \frac{1}{2}\int \frac{du}{u}. $$
The integral of $\dfrac{1}{u}$ is:
$$ I = \frac{1}{2}\ln|u| + C. $$
Step 3 · Substitute back
Substituting back $u = x^2+6x-5$, we get:
$$ I = \frac{1}{2}\ln\left|x^2+6x-5\right| + C. $$
Summary Table
| Quantity | Value |
|---|---|
| Factored denominator | $(x+5)(x-1)$ |
| Substitution | $u = x^2+6x-5$ |
| Antiderivative | $\tfrac12\ln|u|$ |
Additional Notes
- The domain of $I$ requires $x \neq -5,\ 1$, where the denominator vanishes.
- Compare with Problem 67: there the numerator matched the derivative exactly, requiring no extra constant; here the numerator is only half of the derivative, which introduces the $\tfrac12$ coefficient in the antiderivative.
- The result can be checked by differentiating $\tfrac12\ln\left|x^2+6x-5\right|$ with the chain rule and confirming it returns $\dfrac{x+3}{x^2+6x-5}$.
Result
The solution to the integral is: