evaluate the integral $int\frac{xdx}{(x^{2}+4)^{4}}$

Explanation:

Step1: Use substitution

Let $u = x^{2}+4$, then $du = 2xdx$, and $xdx=\frac{1}{2}du$.

Step2: Rewrite the integral

The integral $\int\frac{xdx}{(x^{2}+4)^{4}}$ becomes $\frac{1}{2}\int u^{- 4}du$.

Step3: Integrate using power - rule

The power - rule for integration is $\int x^{n}dx=\frac{x^{n + 1}}{n+1}+C$ ($n\neq - 1$). For $\frac{1}{2}\int u^{-4}du$, we have $\frac{1}{2}\times\frac{u^{-4 + 1}}{-4+1}+C=\frac{1}{2}\times\frac{u^{-3}}{-3}+C=-\frac{1}{6u^{3}}+C$.

Step4: Substitute back $u$

Substitute $u = x^{2}+4$ back into the result, we get $-\frac{1}{6(x^{2}+4)^{3}}+C$.

Answer:

$-\frac{1}{6(x^{2}+4)^{3}}+C$