What are $b$ and $c$ if $\int^1_0e^x \left[c\ln(x^2+1)+\frac{bx}{x^2+1} \right]dx=\frac{be}{2}\ln2$ I came across a question today...

Let $F(x) =  e^x \left(c\ln(x^2+1)+\dfrac{bx}{x^2+1} \right)$. If $\displaystyle \int^1_0F(x)\,dx=\dfrac{b e}{2}\ln2$ then the values of $c$ and $b$ can be:
  Options are
  
  
*
  
*$c=1, b=2$  
  
*$c=2, b=3$  
  
*$c=1/2, b=1$  
  
*$c=1/3, b=1/2$
  

I first tried to integrate $F(x)$. I got...$$\int F(x)dx=c e^x\ln(1+x^2)+(b-2c) \int \dfrac{x e^xdx}{x^2+1}.$$
Well ...now what? I have no idea how to solve it now?  Is it even a right way to do such a question?

Notice that integration by parts gives
$$\int_0^1\frac{bxe^{x}dx}{1+x^2}=\frac{be}{2}\ln2-\frac b2\int_0^1e^x
\ln\left(1+x^2\right)dx.$$
Hence your equation is equivalent to
$$\left(c-\frac b2\right)\int_0^1e^x
\ln\left(1+x^2\right)dx=0.$$
I hope the rest is clear.
