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Integration by substitution
Problem: evaluate the integral from 4 to 9 of 1/x1/2+4 dx
A good strategy for integration using a u substitution is to see if it's possible to write your function as a product of some "inner" function and this inner function's derivative.Originally Posted by Kvandesterren
Here we have. Since we have
inside our function, we will probably need to have this in our "inner" function that we make u. Which means we will need its derivative,
as a factor in our integrand. So by manipulating the integrand slightly...
and now it's relatively easy to see that a substitution ofwill work nicely. Note that when
and when
, and also that
, then the integral becomes
![\displaystyle \begin{align*} 2\int_4^9{\left(\frac{\sqrt{x}}{\sqrt{x} + 4}\right)\frac{1}{2\sqrt{x}}\,dx} &= 2\int_6^7{\frac{u - 4}{u}\,du} \\ &= 2\int_6^7{1 - \frac{4}{u}\,du} \\ &= 2\left[u - 4\ln{|u|}\right]_6^7 \\ &= 2\left[\left(7 - 4\ln{7}\right) - \left(6 - 4\ln{6}\right)\right] \\ &= 2\left(1 + 4\ln{6} - 4\ln{7}\right) \\ &= 2 + 8\ln{6} - 8\ln{7}\end{align*}](http://latex.codecogs.com/png.latex?\displaystyle \begin{align*} 2\int_4^9{\left(\frac{\sqrt{x}}{\sqrt{x} + 4}\right)\frac{1}{2\sqrt{x}}\,dx} &= 2\int_6^7{\frac{u - 4}{u}\,du} \\ &= 2\int_6^7{1 - \frac{4}{u}\,du} \\ &= 2\left[u - 4\ln{|u|}\right]_6^7 \\ &= 2\left[\left(7 - 4\ln{7}\right) - \left(6 - 4\ln{6}\right)\right] \\ &= 2\left(1 + 4\ln{6} - 4\ln{7}\right) \\ &= 2 + 8\ln{6} - 8\ln{7}\end{align*})
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