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DETAILS SCALC9 4.5.003.MI.SA.
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Tutorial Exercise
Evaluate the integral by making the given substitution.
\[
\int x^{2} \sqrt{x^{3}+24} d x, \quad u=x^{3}+24
\]
Step 1
We know that if $u=f(x)$, then $d u=f^{\prime}(x) d x$. Therefore, if $u=x^{3}+24$, then $d u=$ $d x$
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Step 1 ：Given the integral \(\int x^{2} \sqrt{x^{3}+24} dx\) and the substitution \(u=x^{3}+24\).

Step 2 ：Differentiate \(u=x^{3}+24\) with respect to \(x\) to find \(du\). This gives \(du=3x^{2}dx\).

Step 3 ：Rearrange the equation \(du=3x^{2}dx\) to solve for \(dx\). This gives \(dx=\frac{du}{3x^{2}}\).

Step 4 ：Substitute \(u\) and \(dx\) into the integral. This gives \(\int x^{2} \sqrt{u} \frac{du}{3x^{2}}\).

Step 5 ：Simplify the integral to \(\int \frac{1}{3} \sqrt{u} du\).

Step 6 ：Evaluate the integral to find \(\frac{2}{9} u^{\frac{3}{2}} + C\), where \(C\) is the constant of integration.

Step 7 ：Substitute \(x^{3}+24\) back in for \(u\) to find the final answer. This gives \(\boxed{\frac{2}{9} (x^{3}+24)^{\frac{3}{2}} + C}\).