Question

evaluate (be sure to check by differentiating!)
intleft(8+x^{4}
ight)^{5} 4 x^{3} d x
intleft(8+x^{4}
ight)^{5} 4 x^{3} d x=square
(type an exact answer. use parentheses to clearly denote the argument of each function )

Explanation:

Step1: Substitution

Let \(u = 8 + x^{4}\), then \(du=4x^{3}dx\).
The integral \(\int(8 + x^{4})^{5}4x^{3}dx\) becomes \(\int u^{5}du\).

Step2: Integrate

Using the power - rule \(\int u^{n}du=\frac{u^{n + 1}}{n+1}+C\) (\(n
eq - 1\)), for \(n = 5\), we have \(\int u^{5}du=\frac{u^{6}}{6}+C\).

Step3: Back - substitution

Substitute \(u = 8 + x^{4}\) back into the result. So \(\frac{u^{6}}{6}+C=\frac{(8 + x^{4})^{6}}{6}+C\).

Step4: Check by differentiating

Differentiate \(y=\frac{(8 + x^{4})^{6}}{6}+C\) using the chain rule.
Let \(t = 8+x^{4}\), then \(y=\frac{t^{6}}{6}+C\). \(\frac{dy}{dt}=t^{5}\) and \(\frac{dt}{dx}=4x^{3}\).
By the chain rule \(\frac{dy}{dx}=\frac{dy}{dt}\cdot\frac{dt}{dx}=(8 + x^{4})^{5}\cdot4x^{3}\), which is the integrand.

Answer:

\(\frac{(8 + x^{4})^{6}}{6}+C\)