1. Use partial fractions to solve
\(\displaystyle \int\frac{x-7}{x^{2}(x+2)}dx\)
Expand into partial fractions: \(\displaystyle \frac{-9}{4}\int\frac{1}{(x+2)}dx+\frac{9}{4}\int\frac{1}{x}dx-\frac{7}{2}\int\frac{1}{x^{2}}dx\)
2. Solve the following integral.
\(\displaystyle 5\int\frac{1}{\sqrt{x^{2} - x - 2}}dx\)
[/quote]
There are various ways to proceed. One could factor what is inside the radical and make a u substitution. Or use completing the square. Tri sub is another option.
\(\displaystyle 5\int\frac{1}{\sqrt{(x-2)(x+1)}}dx\)
Complete the square and get \(\displaystyle 5\int\frac{1}{\sqrt{(x-\frac{1}{2})^{2}-\frac{9}{4}}}dx\)
Let \(\displaystyle u=x-\frac{1}{2}, \;\ du=dx\) and get:
\(\displaystyle \int\frac{10}{\sqrt{4u^{2}-9}}du\)
Now, a simple trig sub will transform it into terms of secant and you have it.
Let \(\displaystyle u=\frac{3}{2}sec{\theta}, \;\ du=\frac{3}{2}sec{\theta}tan{\theta}d{\theta}\)
Just one of many ways to skin an integral.
Another way would be as I mentioned before.
\(\displaystyle 5\int\frac{1}{\sqrt{(x-2)(x+1)}}dx\)
Let \(\displaystyle u=x+1, \;\ u-3=x-2, \;\ du=dx\)
\(\displaystyle 5\int\frac{1}{\sqrt{(u-3)u}}du\)
