find the indefinite integral. (remember the constant of integration.)\n∫(6x^5 + 3)/(x^6 + 3x)^2 dx

Explanation:

Step1: Use substitution

Let $u = x^{6}+3x$. Then $du=(6x^{5}+3)dx$.

Step2: Rewrite the integral

The integral $\int\frac{6x^{5}+3}{(x^{6}+3x)^{2}}dx$ becomes $\int\frac{du}{u^{2}}$.

Step3: Integrate $\frac{1}{u^{2}}$

We know that $\int u^{n}du=\frac{u^{n + 1}}{n+1}+C$ for $n\neq - 1$. Here $n=-2$, so $\int\frac{du}{u^{2}}=\int u^{-2}du=\frac{u^{-2 + 1}}{-2+1}+C=-\frac{1}{u}+C$.

Step4: Substitute back $u$

Substituting $u = x^{6}+3x$ back, we get $-\frac{1}{x^{6}+3x}+C$.

Answer:

$-\frac{1}{x^{6}+3x}+C$