Question

evaluate the integral
int x ^ { 5 } left( x ^ { 6 } - 8
ight) ^ { 8 } d x
by making the substitution ( u = x ^ { 6 } - 8 ).

+ c
note: your answer should be in terms of ( x ) and not ( u ).
question help: video

Explanation:

Step1: Find the derivative of \(u\)

Given \(u = x^{6}-8\), then \(du=6x^{5}dx\), so \(x^{5}dx=\frac{1}{6}du\).

Step2: Substitute into the integral

The integral \(\int x^{5}(x^{6} - 8)^{8}dx\) becomes \(\int u^{8}\cdot\frac{1}{6}du\).

Step3: Integrate with respect to \(u\)

Using the power - rule \(\int x^{n}dx=\frac{x^{n + 1}}{n+1}+C(n
eq - 1)\), we have \(\frac{1}{6}\int u^{8}du=\frac{1}{6}\cdot\frac{u^{9}}{9}+C=\frac{u^{9}}{54}+C\).

Step4: Substitute back \(u=x^{6}-8\)

We get \(\frac{(x^{6}-8)^{9}}{54}+C\).

Answer:

\(\frac{(x^{6}-8)^{9}}{54}\)