Covers more complex u-substitution problems where it's necessary to algebraically solve for the original variable (e.g., x) in terms of u and substitute it back into the integral.
In what situation would you need to solve for \(x\) in terms of \(u\) during a u-substitution?
This technique is used when, after substituting for \(u\) and \(du\), there is still a term with the original variable \(x\) remaining in the integrand that wasn't eliminated by the \(du\) substitution.
Consider the integral \(\int x \sqrt{x-3} \, dx\). If you choose \(u = x-3\), what are \(du\) and \(x\)?
If \(u = x-3\), then \(du = dx\). By rearranging the substitution equation, we find that \(x = u+3\).
Using \(u = x-3\), \(du=dx\), and \(x=u+3\), rewrite the integral \(\int x \sqrt{x-3} \, dx\) entirely in terms of \(u\).
Substitute each part: \(x\) becomes \(u+3\) and \(\sqrt{x-3}\) becomes \(\sqrt{u}\). The integral becomes \(\int (u+3) \sqrt{u} \, du\).
How do you proceed to integrate \(\int (u+3) \sqrt{u} \, du\)?
First, distribute \(\sqrt{u}\) (which is \(u^{1/2}\)) across the parentheses: \(\int (u \cdot u^{1/2} + 3u^{1/2}) \, du = \int (u^{3/2} + 3u^{1/2}) \, du\). Then integrate term by term.
Find the final solution for \(\int x \sqrt{x-3} \, dx\).
For \(\int x^3 (x^2+1)^{1/2} \, dx\), let \(u=x^2+1\). What are \(x^2\) and \(du\)?
From the substitution, \(x^2 = u-1\). The differential is \(du = 2x \, dx\).
How do you handle the \(x^3\) term in \(\int x^3 (x^2+1)^{1/2} \, dx\) using \(u=x^2+1\)?
Split \(x^3\) into \(x^2 \cdot x\). The integral is \(\int x^2 \sqrt{x^2+1} \cdot x \, dx\). Now you can substitute \(x^2 = u-1\), \(\sqrt{x^2+1} = \sqrt{u}\), and \(x \, dx = \frac{1}{2}du\).
Rewrite \(\int x^3 (x^2+1)^{1/2} \, dx\) completely in terms of \(u\), given \(u=x^2+1\).
