Question

Can ∫ (x^6 +8)^2 dx be integrated with u = x^6+8? Explain

Answer 1

$u=x^6+8$ is a differentiable function. Its rang is $(-\infin, \infin).$

The fuction$f(x)=(x^6+8)^2$ is continuous on $(-\infin, \infin).$

Then we can use the Substitution Rule and

$$u=x^6+8, du=6x^5 dx$$

$$\int(x^6+8)^2 (6x^5)dx=\int u^2du=\dfrac{u^3}{3}+C$$

$$=\dfrac{(x^6+8)^3}{3}+C$$

In our case we have

$$\int(x^6+8)^2 dx$$

instead of

$$\int(x^6+8)^2 (6x^5)dx$$

Therefore it is not useful to integrate

$$\int(x^6+8)^2 dx$$

with $u-$ substitution $u=x^6+8.$

$$\int(x^6+8)^2 dx=\int(x^{12}+16x^6+64) dx$$

$$=\dfrac{x^{13}}{13}+\dfrac{16x^7}{7}+64x+C$$