The definite integral of $\int_0^u \ln(ax)/\sqrt{u-x}\,dx$

Question

I need to solve the integral:

$$\int_0^u \frac{\ln(ax)}{\sqrt{u-x}}\,dx$$

Please help me out. Thanks in advance.

Answer 1

Let us assume $$f(a) =\int_0^u \frac{\ln (ax)}{\sqrt{u-x}} \rm dx $$ Now we've $$f'(a)= \int_0^u \frac{1}{ax} \cdot \frac{ x}{\sqrt{u-x}} \rm dx \implies f'(a) = \frac 1a \int_0^u \frac{1}{\sqrt{u-x}} \rm dx$$

$$f'(a) =-\frac{2 \sqrt{u-x}}{a} \Bigg |_0^u=\frac{2\sqrt u} a$$

$$\implies f(a) = \int \frac{2\sqrt u}{a} \rm da = 2\sqrt u (\ln (a)+C)$$

You can find $C$ by using some intial conditions. Note that $C$ can depend on $u$ only, not on $a$)