Question

integration of under root of a+x/x

Answer 1

We have to find the indefinite integral $\int \sqrt{\frac{a+x}{x}} \,dx$.

Let us make the substitution, $\frac{a+x}{x}=t^2,x=\frac{a}{t^2-1} ,dx=-\frac{2at}{(t^2-1)^2}dt$.

Under the above substitution the integral becomes,

$\int \sqrt{\frac{a+x}{x}} \,dx =\int \sqrt{t^2} \frac{2at}{(t^2-1)^2}dt \\ \int \sqrt{\frac{a+x}{x}} \,dx =\int at \frac{2t}{(t^2-1)^2}dt$

Use integration by parts,

$\int \sqrt{\frac{a+x}{x}} \,dx =-at \frac{1}{(t^2-1)}dt+a\int \frac{1}{(t^2-1)}dt \\ \int \sqrt{\frac{a+x}{x}} \,dx =- \frac{at}{(t^2-1)}dt+\frac{a}{2} \int (\frac{1}{t-1}-\frac{1}{t+1}) dt\\ \int \sqrt{\frac{a+x}{x}} \,dx =- \frac{at}{(t^2-1)}dt+\frac{a}{2}\ln |\frac{t-1}{t+1}|+C$

Back substituting,

$\int \sqrt{\frac{a+x}{x}} \,dx =- \frac{a\sqrt{\frac{a+x}{x}}}{\frac{a+x}{x}-1}+\frac{a}{2}\ln |\frac{\sqrt{\frac{a+x}{x}}-1}{\sqrt{\frac{a+x}{x}}+1}|+C \\ \int \sqrt{\frac{a+x}{x}} \,dx =-x\sqrt{\frac{a+x}{x}}++\frac{a}{2}\ln |\frac{\sqrt{a+x}-\sqrt{x}}{\sqrt{\sqrt{a+x}+\sqrt{x}}}|+C$