Question

illustrate and check that your answer is reasonable by graphing both the function f and its antiderivative f (take c = 0). $int x(x^{2}-3)^{3}dx=\frac{(x^{2}-3)^{4}}{8}+c$

Explanation:

Step1: Use substitution method

Let \(u = x^{2}-3\), then \(du=2x dx\), and \(x dx=\frac{1}{2}du\). The integral \(\int x(x^{2}-3)^{3}dx\) becomes \(\frac{1}{2}\int u^{3}du\).

Step2: Integrate \(u^{3}\)

According to the power - rule for integration \(\int x^{n}dx=\frac{x^{n + 1}}{n+1}+C\) (\(n
eq - 1\)), for \(\frac{1}{2}\int u^{3}du\), we have \(\frac{1}{2}\times\frac{u^{4}}{4}+C=\frac{u^{4}}{8}+C\).

Step3: Substitute back \(u\)

Substitute \(u = x^{2}-3\) back into the result, we get \(\frac{(x^{2}-3)^{4}}{8}+C\).

Answer:

\(\frac{(x^{2}-3)^{4}}{8}+C\)