Michael Bowen's VC Course Pages
Answers
Math V21B Section 7.1
Introduction
Please try this problem on your own before looking at my solution. There may be other ways to solve this; I make no claim that this method is the quickest, prettiest, or most efficient. Also, I did not look at the student solutions manual, so this approach may or may not match what they did.
Problem 21
The original problem is $$I = \int {\frac{{x{e^{2x}}}}{{{{\left( {1 + 2x} \right)}^2}}}dx} $$
As this function does not clearly adhere to the "LIPTE" mnemonic, we may need to experiment with techniques. After some trial and error, we may discover that integration by parts, starting with $u = x{e^{2x}}$, has the potential to be helpful. With this as the initial choice, the remaining substitutions become $$dv = {(1 + 2x)^{ - 2}}dx$$ $$du = \left( {{e^{2x}} + 2x{e^{2x}}} \right)dx = {e^{2x}}\left( {1 + 2x} \right)$$ (note use of the product rule for $du$), and $$v = \frac{{ - 1}}{{2\left( {1 + 2x} \right)}}$$ How to obtain the latter result from $dv$ may not be obvious; start with $$v = \int {dv} = \int {{{\left( {1 + 2x} \right)}^{ - 2}}dx} $$ and apply the $w$-substitution $$w = \left( {1 + 2x} \right);dw = 2dx;dx = \frac{1}{2}dw$$ to obtain $$v = \frac{1}{2}\int {{w^{ - 2}}dw} = \frac{1}{2}\frac{{{w^{ - 1}}}}{{ - 1}} = \frac{{ - 1}}{{2w}} = \frac{{ - 1}}{{2\left( {1 + 2x} \right)}}$$ Using the integration-by-parts mechanism yields $$I = x{e^{2x}}\left[ {\frac{{ - 1}}{{2\left( {1 + 2x} \right)}}} \right] - \int {\frac{{ - 1}}{{2\left( {1 + 2x} \right)}}} {e^{2x}}\left( {1 + 2x} \right)dx = x{e^{2x}}\left[ {\frac{{ - 1}}{{2\left( {1 + 2x} \right)}}} \right] + \frac{1}{2}\int {{e^{2x}}dx} $$ after canceling some negative signs and factors of $\left( {1 + 2x} \right)$ in the integral.
The remaining integral can be evaluated using a simple $w$-substitution ($w = 2x$), giving $$I = x{e^{2x}}\left[ {\frac{{ - 1}}{{2\left( {1 + 2x} \right)}}} \right] + \frac{1}{4}{e^{2x}} + C$$ which would be the minimally acceptable final answer on an exam. We can make this prettier, however, by applying a bit of algebraic simplification, starting by factoring out ${e^{2x}}$ and then changing both fractions inside the brackets to have a common denominator: $$I = {e^{2x}}\left[ {\frac{{ - x}}{{2\left( {1 + 2x} \right)}} + \frac{1}{4}} \right] + C = {e^{2x}}\left[ {\frac{{ - 2x}}{{4\left( {1 + 2x} \right)}} + \frac{{1\left( {1 + 2x} \right)}}{{4\left( {1 + 2x} \right)}}} \right] + C = {e^{2x}}\left[ {\frac{{ - 2x + 1 + 2x}}{{4\left( {1 + 2x} \right)}}} \right] + C$$
This simplifies nicely to $$\boxed{\displaystyle{I = \frac{{{e^{2x}}}}{{4\left( {1 + 2x} \right)}} + C}}$$
