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?

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


Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.