Finding area of a region in terms of a variable

Question

The curves $y=mx$ and $y=\frac{x}{x^2+1}$ enclose a region for $0<m<1$. Find the area of a region in terms of only m. The integral I set up is the following $$A=2\int_{a}^{0}\left(mx-\frac{x}{x^2+1}\right)dx=ma^2+\ln|a^2+1|$$ The answer I sholuld get is $A=m-\ln|m-1|$.How do I get there ? Can I express $a$ in terms of $m$ ?

Answer 1

The points of intersection are the roots of \begin{align} \frac{x}{x^2+1}-mx&=0 , \end{align}

\begin{align} x&=0,\ \pm \sqrt{\frac1m-1} . \end{align}

Since $m<1$, the line $y=mx$ is below the curve $\frac{x}{x^2+1}$ for $x\in(0,\sqrt{\frac1m-1})$, hence the area is

\begin{align} A&= 2\,\int_0^{\sqrt{\frac1m-1}} \left( \frac{x}{x^2+1}-mx \right)\,dx \\ &=\left. 2\,\left(\tfrac12\ln(x^2+1) -\tfrac12\,m x^2 \right)\right|_{0}^{\sqrt{\frac1m-1}} \\ &=m-1-\ln m . \end{align}

Answer 2

Solvingt the equation $$mx=\frac{x}{x^2+1}$$ we get $$x=0$$ or $$x=\pm \frac{\sqrt{1-m}}{m}$$ so we get $$A=2\int_{0}^{\sqrt{\frac{1-m}{m}}}mx-\frac{x}{1+x^2}dx$$