Evaluate the integral: I = integral dx/((x²-4x+4)

Evaluate the integral: $$ I = \int \frac{dx}{\left(x^2-4x+4\right)\left(x^2-4x+5\right)}. $$

Step 1 · Recognize the structure

$$ x^2-4x+4=(x-2)^2, \qquad x^2-4x+5=(x-2)^2+1. $$

Step 2 · Substitute

Let $u=x-2$, $du=dx$:

$$ I = \int \frac{du}{u^2(u^2+1)}. $$

Step 3 · Partial fractions

$$ \frac{1}{u^2(u^2+1)} = \frac Au + \frac{B}{u^2} + \frac{Cu+D}{u^2+1}. $$

Multiplying through and matching coefficients gives $A=0$, $B=1$, $C=0$, $D=-1$.

Step 4 · Integrate and substitute back

$$ \frac{1}{u^2(u^2+1)} = \frac{1}{u^2} - \frac{1}{u^2+1}, \qquad \int \left(\frac{1}{u^2}-\frac{1}{u^2+1}\right) du = -\frac1u - \arctan u + C. $$

With $u=x-2$:

$$ I = -\frac{1}{x-2} - \arctan(x-2) + C. $$

Summary Table

Additional Notes

• The domain excludes $x=2$, where the first factor vanishes.
• Both quadratic factors share the same "core" $(x-2)^2$, differing only by an additive constant — recognizing this before expanding either quadratic makes the substitution $u=x-2$ the obvious next step, avoiding a messier direct partial-fraction setup in $x$.
• The result can be checked by differentiating $-\tfrac{1}{x-2}-\arctan(x-2)$: this gives $\tfrac{1}{(x-2)^2}-\tfrac{1}{(x-2)^2+1}=\tfrac{[(x-2)^2+1]-(x-2)^2}{(x-2)^2[(x-2)^2+1]}=\tfrac{1}{(x-2)^2[(x-2)^2+1]}$, matching the integrand since $(x-2)^2=x^2-4x+4$ and $(x-2)^2+1=x^2-4x+5$.