@Axel Vogt 

Thanks for your note. But unfortunately, I didn't understand. 

@acer Thanks

@acer 

Dear acer

Why your second suggestion does not work for a similar equation? Please guide me.

 r.mw

@acer 

That's great :)

Thanks a lot

@acer 

Thank you so much

@Mariusz Iwaniuk 

Dear Mariusz

thanks for the code but the denaminator is ((sqrt(a*e^(-2*x)+a))-sqrt(a))). not ((sqrt(a*e^(-2*x)))-sqrt(a))). that make it imposible to solve

Dear Mariusz

thanks for the code but the denaminator is ((sqrt(a*e^(-2*x)+a))-sqrt(a))). not ((sqrt(a*e^(-2*x)))-sqrt(a))). that make it imposible to solve

@golnaz 

thank you for the answe but my integral is indefinite and i want to find an answer i terms of x

@Axel Vogt 

sorry but i dont understand how you write my integral in this form

@Axel Vogt 

Dear Axel Vogt

Sorry, but there was a mistake in my question. I want to solve this:

int((ln(x)/x^2-1/x)^(1/2),x)

Albeit, I have tried to simplify the main integral I want to solve that is:

int((ln(x)^(a-1)/x^2-1/x)^(1/2),x)

This is more complicated. Thanks a lot

@golnaz 

int((ln(x)/x^2-1/x)^1/2,x)

int(sqrt((lnx)^(l-1)/x^2-1/x))

@Preben Alsholm Let's me explain my question exactly. I have:

f(t)=2*beta(t)

and

beta=(t^(1/3);1/5,1/2) 

Now, I want to plot a function as

V = beta^(-1)(f/2)

I want to plot V in terms of "f", not "t"

@Preben Alsholm Thanks a lot for your complete answer. I did not understand again :)

I don't understand how we apply the inverse. However, thank you.

@Preben Alsholm 

Dear Preben. You are right. I mean an incomplete beta function and I mean the inverse. I plotted my fugure. But one more question. I did not understand how you apply the inverse. In the plot command

plot([eval(b,t=f/2),f,f=0..2]);

we want to eval b for t=f/2 and b is itself the incomplete beta function

b:=eval(B,{x=sqrt(t),a=1/5,b=1/2});

and not its inverse. Am I wrong? Could you please explain more?