Find the integral of dx / ((x − 1) sqrt(−x² + 2x + 3)).

Step-by-step solution

1. 1

   Complete the square under the radical. The radicand is

   $-x^{2}+2x+3=4-(x-1)^{2}$

   so the integral is real only for $|x-1|<2$, i.e. $-1<x<3$, and $x\neq 1$. The appearance of $(x-1)$ both outside and inside is the clue to the right substitution.

2. 2

   Use the reciprocal substitution. For an integrand of the form $\dfrac{1}{(x-a)\sqrt{\text{quadratic}}}$, set $x-a=\dfrac1t$. Here

   $x-1=\frac1t\quad\Longrightarrow\quad dx=-\frac{dt}{t^{2}}$

   (Note the minus sign: differentiating $1/t$ gives $-1/t^{2}$. Dropping it flips the sign of the whole answer.)

3. 3

   Rewrite the radical. With $(x-1)^{2}=\tfrac{1}{t^{2}}$,

   $\sqrt{4-(x-1)^{2}}=\sqrt{4-\frac{1}{t^{2}}}=\frac{\sqrt{4t^{2}-1}}{|t|}$

4. 4

   Assemble the transformed integral.

   $I=\int\frac{-\dfrac{dt}{t^{2}}}{\dfrac1t\cdot\dfrac{\sqrt{4t^{2}-1}}{|t|}}=-\int\frac{|t|}{t}\cdot\frac{dt}{\sqrt{4t^{2}-1}}=-\operatorname{sgn}(t)\int\frac{dt}{\sqrt{4t^{2}-1}}$

   The messy $t^{2}$ factors cancel completely — the point of the substitution.

5. 5

   Integrate the standard form and substitute back. Since $\displaystyle\int\frac{dt}{\sqrt{4t^{2}-1}}=\frac12\ln\left|2t+\sqrt{4t^{2}-1}\right|+C$, and $2t+\sqrt{4t^{2}-1}=\dfrac{2\operatorname{sgn}(x-1)+\sqrt{-x^{2}+2x+3}}{|x-1|}$, the sign factor combines with the logarithm to give the single formula

   $\boxed{I=-\frac12\ln\left|\frac{2+\sqrt{-x^{2}+2x+3}}{x-1}\right|+C}$

   valid on both $-1<x<1$ and $1<x<3$.

6. 6

   Verify by differentiating and by a numerical check. Differentiating the answer returns $\dfrac{1}{(x-1)\sqrt{-x^{2}+2x+3}}$. Numerically, $\displaystyle\int_{1.5}^{2.5}\frac{dx}{(x-1)\sqrt{-x^{2}+2x+3}}=0.634036$, and the closed form gives $F(2.5)-F(1.5)=0.634036$ ✓; on the other branch, $\displaystyle\int_{0.2}^{0.8}=-0.713212$ and the formula also gives $-0.713212$ ✓.

7. 7

   Watch two sign traps. Because $(2+\sqrt{\cdot})(2-\sqrt{\cdot})=(x-1)^{2}$, the alternative form $\tfrac12\ln\left|\tfrac{2-\sqrt{-x^{2}+2x+3}}{x-1}\right|$ differs only by a constant and is equally valid; but the version without the leading minus sign is wrong on the branch $x<1$, and using $dx=+dt/t^{2}$ instead of $-dt/t^{2}$ negates the whole antiderivative.