Find the indefinite integral of (x - 1) / (x^2 - 2x + 3)^2 with respect to x.

Step-by-step solution

1. 1

   Look at the derivative of the denominator's base. Before reaching for partial fractions, differentiate what is being squared:

   $\frac{d}{dx}\left(x^2-2x+3\right)=2x-2=2(x-1)$

   The numerator $x-1$ is exactly half of this. Whenever the numerator is a constant multiple of the inner derivative, substitution finishes the integral in one line — no completing the square, no arctangent.

2. 2

   Substitute. Let

   $u=x^2-2x+3,\qquad du=2(x-1)\,dx\quad\Longrightarrow\quad (x-1)\,dx=\tfrac12\,du$

   The whole $(x-1)\,dx$ block is replaced at once, which is why no leftover $x$ survives.

3. 3

   Rewrite the integral in $u$.

   $\int\frac{x-1}{(x^2-2x+3)^2}\,dx=\frac12\int \frac{du}{u^{2}}=\frac12\int u^{-2}\,du$

4. 4

   Integrate the power. With the power rule $\int u^{n}\,du=\frac{u^{n+1}}{n+1}$ for $n=-2$:

   $\frac12\int u^{-2}\,du=\frac12\cdot\frac{u^{-1}}{-1}=-\frac{1}{2u}$

5. 5

   Substitute back and add the constant.

   $\int \frac{x-1}{\left(x^2-2x+3\right)^2}\,dx=-\frac{1}{2\left(x^2-2x+3\right)}+C$

   The denominator $x^2-2x+3=(x-1)^2+2$ is never zero, so this antiderivative is valid on all of $\mathbb{R}$.

6. 6

   Differentiate to confirm. With $F(x)=-\tfrac12 u^{-1}$,

   $F'(x)=\tfrac12 u^{-2}u'=\frac{2x-2}{2\left(x^2-2x+3\right)^2}=\frac{x-1}{\left(x^2-2x+3\right)^2}\;\checkmark$

   A numerical check at $x=2.5$ gives $F'(2.5)=0.08304498$, matching the integrand exactly.