The question is:
The curve $(C')$ in the above figure is the represantative curve of the function $f$ defined over $]0,+\infty[$ by $f(x)=\frac{a \ln x}{b+x}$.
Find $a$ and $b$ knowing that $x=0$ is an asymptote to $(C')$ and that the area of the domain limited by $(C')$, the $x$-axis and the straight line of equation $x=2$ is $A=\frac{(\ln 2)^2}{2}$
I found 1 of the equations from the fact that $f'(1)=1$
But I tried to use the area and the curve and I didn't know how to integrate this function $f(x)$ from $x=1$ to $x=2$
And the asymptotes also didn't help and neither did the fact that $f(1)=0$
Can anyone help with the integral please? Or give another possible solution?
Resume:
The function $g(x)=\frac{\ln x}{x}$ satisfies all constraints, because $\int_1^2 \frac{\ln x}{x} dx =\frac 12 (\ln 2)^2=\mathcal{A}.$ Here $b=0, a=1.$
But $$\frac{a\ln x}{b+x}-\frac{\ln x}{x}=\frac{(x-1)(a-1)\ln x}{x(b+x)}>0 \quad\text{for}\; x\in (1,2]\quad\text{and}\quad \;a=b+1> 1.$$
Therefore we have $$\int_1^2 \frac{a\ln x}{b+x} dx >\frac 12 (\ln 2)^2 \quad \text{if}\quad b>0.$$ The only convenient $b$ is $0$, and $a=1.$