Question

evaluate (be sure to check by differentiating!)
intleft(x^{9}+x^{7}+x^{5}
ight)^{8}left(9 x^{8}+7 x^{6}+5 x^{4}
ight) d x
intleft(x^{9}+x^{7}+x^{5}
ight)^{8}left(9 x^{8}+7 x^{6}+5 x^{4}
ight) d x= (type an exact answer use parentheses to clearly denote the argument of each function )

Explanation:

Step1: Substitute \(u = x^{9}+x^{7}+x^{5}\)

Differentiate \(u\) with respect to \(x\): \(du=(9x^{8}+7x^{6}+5x^{4})dx\)

Step2: Rewrite the integral

The integral \(\int(x^{9}+x^{7}+x^{5})^{8}(9x^{8}+7x^{6}+5x^{4})dx\) becomes \(\int u^{8}du\)

Step3: Integrate \(\int u^{8}du\)

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

Step4: Substitute back \(u = x^{9}+x^{7}+x^{5}\)

The result is \(\frac{(x^{9}+x^{7}+x^{5})^{9}}{9}+C\)

Step5: Check by differentiating

Differentiate \(y=\frac{(x^{9}+x^{7}+x^{5})^{9}}{9}+C\) using the chain rule. Let \(t=x^{9}+x^{7}+x^{5}\), then \(y=\frac{t^{9}}{9}+C\). \(\frac{dy}{dt}=t^{8}\) and \(\frac{dt}{dx}=9x^{8}+7x^{6}+5x^{4}\). By the chain rule \(\frac{dy}{dx}=\frac{dy}{dt}\cdot\frac{dt}{dx}=(x^{9}+x^{7}+x^{5})^{8}(9x^{8}+7x^{6}+5x^{4})\)

Answer:

\(\frac{(x^{9}+x^{7}+x^{5})^{9}}{9}+C\)