Evaluate the integral (for x>1): I = integral
Evaluate the integral (for $x>1$): $$ I = \int \frac{dx}{\sqrt{(x-1)^2(x+2)}}. $$
Explanation of Method
For $x>1$, $\sqrt{(x-1)^2(x+2)}=(x-1)\sqrt{x+2}$, and the substitution $u=\sqrt{x+2}$ reduces the integral to a rational function of $u$ with a difference-of-squares denominator, handled by partial fractions.
- Simplify the radical for $x>1$.
- Substitute $u=\sqrt{x+2}$.
- Apply partial fractions.
- Back-substitute.
Step 1 · Simplify
For $x>1$, $|x-1|=x-1$, so:
$$ I = \int\frac{dx}{(x-1)\sqrt{x+2}}. $$
Step 2 · Substitute
Let $u=\sqrt{x+2}$, so $x=u^2-2$, $dx=2u\,du$, and $x-1=u^2-3$:
$$ I = \int\frac{2u\,du}{(u^2-3)u} = \int\frac{2\,du}{u^2-3}. $$
Step 3 · Partial fractions
$$ \frac2{u^2-3} = \frac1{\sqrt3}\left(\frac1{u-\sqrt3}-\frac1{u+\sqrt3}\right). $$
$$ I = \frac1{\sqrt3}\ln\left|\frac{u-\sqrt3}{u+\sqrt3}\right| + C. $$
Step 4 · Back-substitute
With $u=\sqrt{x+2}$, and rationalizing by multiplying inside the log by $(\sqrt{x+2}-\sqrt3)/(\sqrt{x+2}-\sqrt3)$ so the denominator becomes $(x+2)-3=x-1$:
$$ I = \frac1{\sqrt3}\ln\left|\frac{\left(\sqrt{x+2}-\sqrt3\right)^2}{x-1}\right| + C. $$
Summary Table
| Substitution | Effect |
|---|---|
| $u=\sqrt{x+2}$ | reduces to $\int2\,du/(u^2-3)$ |
Additional Notes
- The domain here is $x>1$; a symmetric derivation handles $-2<x<1$.
- The result was verified by differentiation at $x=2$ and $x=5$: the derivative matches the integrand $1/\sqrt{(x-1)^2(x+2)}$ exactly at both points.
Result
The solution to the integral is: