Evaluate the integral: I = integral (x+1)/(sqrt(x² + 2x
Evaluate the integral: $$ I = \int \frac{x+1}{\sqrt{x^2 + 2x - 9}} \, dx. $$
Explanation of Method
Completing the square turns the quadratic under the root into a shifted difference of squares, $(x+1)^2-10$. After the shift $u=x+1$, the numerator matches the derivative pattern needed for the standard result $\int \frac{u}{\sqrt{u^2-a^2}}\,du=\sqrt{u^2-a^2}+C$.
- Complete the square under the radical.
- Substitute $u$ for the shifted variable.
- Recognize the standard form $\int \frac{u}{\sqrt{u^2-a^2}}\,du$.
- Substitute back and add the constant of integration.
Step 1 · Complete the square
The expression inside the square root in the denominator is $x^2+2x-9$. To simplify this, we complete the square on $x^2+2x$:
$$ x^2 + 2x = (x+1)^2 - 1. $$
Thus, the original quadratic becomes:
$$ x^2+2x-9 = (x+1)^2 - 10. $$
Therefore, the integral becomes:
$$ I = \int \frac{x+1}{\sqrt{(x+1)^2-10}}\,dx. $$
Step 2 · Substitution
Let $u = x+1$. This gives $du = dx$, and the integral becomes:
$$ I = \int \frac{u}{\sqrt{u^2-10}}\,du. $$
Step 3 · Recognizing the standard form
We recognize that this is now a standard form integral of the type $\int \frac{u}{\sqrt{u^2-a^2}}\,du$, which has the known solution:
$$ \int \frac{u}{\sqrt{u^2-a^2}}\,du = \sqrt{u^2-a^2} + C. $$
Here, $a^2=10$, so the solution is:
$$ \sqrt{u^2-10} + C. $$
Step 4 · Substituting back $u = x + 1$
Now, substituting $u = x+1$ back into the solution:
$$ \sqrt{(x+1)^2-10} + C. $$
This simplifies to:
$$ \sqrt{x^2+2x-9} + C. $$
Summary Table
| Quantity | Value |
|---|---|
| Completed square | $(x+1)^2-10$ |
| Substitution | $u = x+1$ |
| Antiderivative | $\sqrt{u^2-10}$ |
Additional Notes
- The domain of $I$ requires $(x+1)^2 > 10$, i.e. $x < -1-\sqrt{10}$ or $x > -1+\sqrt{10}$, so the expression under the root is positive.
- Completing the square is the standard way to turn a general quadratic $x^2+bx+c$ under a radical into the shifted form $(x+b/2)^2 - (\text{constant})$, which then matches a known standard integral.
- The result can be checked by differentiating $\sqrt{x^2+2x-9}$ with the chain rule and confirming it returns $\dfrac{x+1}{\sqrt{x^2+2x-9}}$.
Result
The solution to the integral is: