Problem 336
Easy Advanced Techniques

Evaluate the integral: I = integral

Evaluate the integral: $$ I = \int \frac{x}{(x-1)^2\sqrt{1+2x-x^2}}\,dx. $$

Explanation of Method

Shifting to $u=x-1$ centers the quadratic as $2-u^2$ and turns the numerator $x=u+1$ into two standard pieces: $\int du/[u\sqrt{2-u^2}]$ and $\int du/[u^2\sqrt{2-u^2}]$, both of which have classical closed forms.

  1. Substitute $u=x-1$.
  2. Split into two standard integrals.
  3. Apply the standard formulas.
  4. Back-substitute.

Step 1 · Substitute

Let $u=x-1$, so $1+2x-x^2=2-u^2$ and $x=u+1$:

$$ I = \int\frac{u+1}{u^2\sqrt{2-u^2}}\,du = \int\frac{du}{u\sqrt{2-u^2}} + \int\frac{du}{u^2\sqrt{2-u^2}}. $$

Step 2 · Apply the standard formulas

Using $\displaystyle\int\frac{du}{u\sqrt{a^2-u^2}}=-\frac1a\ln\left|\frac{a+\sqrt{a^2-u^2}}u\right|$ and $\displaystyle\int\frac{du}{u^2\sqrt{a^2-u^2}}=-\frac{\sqrt{a^2-u^2}}{a^2u}$, with $a=\sqrt2$:

$$ I = -\frac1{\sqrt2}\ln\left|\frac{\sqrt2+\sqrt{2-u^2}}u\right| - \frac{\sqrt{2-u^2}}{2u} + C. $$

Step 3 · Back-substitute

With $u=x-1$, $2-u^2=1+2x-x^2$:

$$ I = -\frac{\sqrt2}2\ln\left|\frac{\sqrt2+\sqrt{1+2x-x^2}}{x-1}\right| - \frac{\sqrt{1+2x-x^2}}{2(x-1)} + C. $$


Summary Table

PieceAntiderivative
$1/(u\sqrt{2-u^2})$$-\tfrac1{\sqrt2}\ln\left|(\sqrt2+\sqrt{2-u^2})/u\right|$
$1/(u^2\sqrt{2-u^2})$$-\sqrt{2-u^2}/(2u)$

Additional Notes

Result

The solution to the integral is:

$$ I = -\frac{\sqrt{1+2x-x^2}}{2(x-1)} - \frac{\sqrt2}2\ln\left|\frac{\sqrt2+\sqrt{1+2x-x^2}}{x-1}\right| + C. $$