Question

$$\int \frac { 4 } { x ^ { 2 } - 8 x + 16 } d x$$

Explanation:

Step1: Complete the square for the denominator

$$x^{2}-8x + 16=(x - 4)^{2}$$
So the integral becomes $\int\frac{4}{(x - 4)^{2}}dx$.

Step2: Use the power - rule for integration

The power - rule for integration is $\int x^{n}dx=\frac{x^{n + 1}}{n+1}+C$ ($n
eq - 1$). Here, if we let $u=x - 4$, then $du=dx$ and our integral is $4\int u^{-2}du$.
Using the power - rule: $4\int u^{-2}du=4\times\frac{u^{-2 + 1}}{-2+1}+C$.

Step3: Simplify the expression

$4\times\frac{u^{-1}}{-1}+C=-\frac{4}{u}+C$.
Substituting back $u = x - 4$, we get $-\frac{4}{x - 4}+C$.

Answer:

$-\frac{4}{x - 4}+C$