# Thread: Substitutions & Partial Fractions

1. ## Substitutions & Partial Fractions

$\displaystyle \int \frac {\sqrt{4-x^2}}{x}dx$

I was thinking along the lines of $\displaystyle x = \cos u$ because $\displaystyle \sin^2x = 1 - \cos^2 x$, but I don't know where to take it.

Also, a partial fractions problem:

$\displaystyle \int \frac {1}{x(x+1)}dx$

These things always turn out to be much simpler than they look.

2. try using x=2cos(u). then you can separate the 4 in the numerator i.e 4-x^2=4(1-cos(u)^2)=4sin(u)^2 which is easy to find the square root of. the rest is pretty straightforward.

The second partial fraction seems relatively simple, all you need to do is let 1/x(x+1)=(A/x)+ (B/(x+1)), then multiplying through by x and then x+1 gives

1=A(x+1) + B(x)

then solve by substituting appropriate values for x, i.e let x=0 and then let x=-1 to find A and B respectively. you will then end up with two separate fractions to integrate with only one function of x in each denominator.