integral of dx/root of c-k/x

Question

∫dx/(c-k/x)^1/2 

Answer 1

First, use the substitution u=1/(c-k/x)^1/2, then du=k dx/(x²(c-k/x)^1/2)= dx (c-u²)^1/2/(2ku), so that

∫dx/(c-k/x)^1/2 = 2k ∫du/(c-u²)².

 

The rest depends on the signs of k and c. Assuming k>0, c>0, use the trigonometric substitution u=√c sec(v), so that

du=√c tan(v)sec(v) dv. Since sec²(v)-1=tan²(v), we have (c-u²)²=c² tan⁴(v) and

∫dx/(c-k/x)^1/2dx = (2k/c^3/2) ∫ sec(v)/tan⁴(v) dv = (2k/c^3/2) ∫ (csc³(v)-csc(v)) dv.

 

The integrals of powers of csc(v) are elementary; you can look them up in a table. For example,

∫ csc(v) dv = ln(tan(v/2)).

 

The final answer, expressed in terms of x, is

k ln(2c^1/2 x(c-k/x)^1/2+2cx-k)/(2c^3/2)+x(c-k/x)^1/2/c.