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Section 7.5 : Integrals Involving Roots

1. Evaluate the integral $$\displaystyle \int{{\frac{7}{{2 + \sqrt {x - 4} }}\,dx}}$$.

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The substitution we’ll use here is,

$u = \sqrt {x - 4}$ Show Step 2

Now we need to get set up for the substitution. In other words, we need so solve for $$x$$ and get $$dx$$.

$x = {u^2} + 4\hspace{0.5in} \Rightarrow \hspace{0.5in}dx = 2u\,du$ Show Step 3

Doing the substitution gives,

$\int{{\frac{7}{{2 + \sqrt {x - 4} }}\,dx}} = \int{{\frac{7}{{2 + u}}\,\left( {2u} \right)du}} = \int{{\frac{{14u}}{{2 + u}}\,du}}$ Show Step 4

This new integral can be done with the substitution $$v = u + 2$$ . Doing this gives,

$\int{{\frac{7}{{2 + \sqrt {x - 4} }}\,dx}} = \int{{\frac{{14\left( {v - 2} \right)}}{v}\,dv}} = \int{{14 - \frac{{28}}{v}\,dv}} = 14v - 28\ln \left| v \right| + c$ Show Step 5

The last step is to now do all the back substitutions to get the final answer.

$\int{{\frac{7}{{2 + \sqrt {x - 4} }}\,dx}} = 14\left( {u + 2} \right) - 28\ln \left| {u + 2} \right| + c = \require{bbox} \bbox[2pt,border:1px solid black]{{14\left( {\sqrt {x - 4} + 2} \right) - 28\ln \left| {\sqrt {x - 4} + 2} \right| + c}}$

Note that we could have avoided the second substitution if we’d used $$u = \sqrt {x - 4} + 2$$ for the original substitution.

This often doesn’t work, but in this case because the only extra term in the denominator was a constant it didn’t change the differential work and so would work pretty easily for this problem.