$int x^{5}(x^{6}-14)^{3}dx=square$

Explanation:

Step1: Use substitution method

Let $u = x^{6}-14$, then $du=6x^{5}dx$, and $x^{5}dx=\frac{1}{6}du$.

Step2: Rewrite the integral

The original integral $\int x^{5}(x^{6} - 14)^{3}dx$ becomes $\frac{1}{6}\int u^{3}du$.

Step3: Integrate $u^{3}$

According to the power - rule of integration $\int x^{n}dx=\frac{x^{n + 1}}{n+1}+C$ ($n\neq - 1$), we have $\frac{1}{6}\times\frac{u^{4}}{4}+C=\frac{u^{4}}{24}+C$.

Step4: Substitute back $u$

Substitute $u = x^{6}-14$ back into the result, we get $\frac{(x^{6}-14)^{4}}{24}+C$.

Answer:

$\frac{(x^{6}-14)^{4}}{24}+C$