# 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

1. $c=1, b=2$
2. $c=2, b=3$
3. $c=1/2, b=1$
4. $c=1/3, b=1/2$

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?

• Substitute the given options and simplify. The substitution which yields $\text{True}$ upon simplification is the right one. – dbanet Mar 5 '16 at 14:05
• @dbanet can i substitute in the integral? – manshu Mar 5 '16 at 14:06
• Sure. But oh well, I didn't notice your primitive. Please notice your integrand is not a differential form. – dbanet Mar 5 '16 at 14:07
• Additionally, since the integration you are originally asked to perform is definite, you should not end up with expressions involving $x$. – dbanet Mar 5 '16 at 14:08
• @dbanet that's the problem with what i tried...it ended up involving x – manshu Mar 5 '16 at 14:09

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.