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The following integral is in question.

$$I(x) =\int_0^x \frac{\ln(1+tx)}{1+t^2}\,dt$$

My attempt is finding $I’(x)$ which is

$$I’(x) = \int_0^x \frac{t}{(1+t^2)(1+tx)}\,dt + \frac{\ln(1+x^2)}{1+x^2} $$ Now we can use partial fraction decompostion for the first integral.

$$\frac{t}{(1+t^2)(1+tx)} = \frac{At +B}{1+t^2}+\frac{C}{1+tx}$$ Solving this gives the following values: $$A = \frac{1}{1+x^2}$$ $$B = \frac{x}{1+x^2}$$ $$C = \frac{-x}{1+x^2}$$

Now we can solve the first integral and we obtain that

$$I’(x) = \frac{1}{2}\frac{\ln(1+x^2)}{1+x^2} + \frac{x\arctan(x)}{1+x^2}$$

Here Intuitively I would get something that is easy to integrate and find the constant by limits or some other obvious way but I get a function that I cannot integrate, can someone help me and give me a hint of what I am doing wrong ? Thanks!!

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  • $\begingroup$ Are you attempting to solve the 2005 Putnam A5 integral? \begin{align} \int_0^1 \frac{\ln(x + 1)}{x^2 + 1}dx \end{align} $\endgroup$
    – NEON
    Commented May 21 at 15:28
  • $\begingroup$ @NEON it was a question on our vector analysis exam, and I got stuck at this part :) $\endgroup$ Commented May 21 at 15:42

1 Answer 1

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We can write $$ 2I'(x) = \frac{1}{1 + x^2}\cdot\ln(1 + x^2) + \tan^{-1}x\cdot \frac{2x}{1+x^2}. $$ Notice any correspondence in the expressions in each term of the sum?

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  • $\begingroup$ Yes, but it feels like some terms should be swapped thats why I am confused, like the derivative of ln() would give the 2x by chain rule but Whereas for arctan we would rather not have the 2x term then its simple substitution. $\endgroup$ Commented May 21 at 15:39
  • $\begingroup$ Now i realised that its actually a derivative of arctanx* ln(1+x^2) $\endgroup$ Commented May 21 at 15:40
  • $\begingroup$ I take it you were looking for something of the form $f'f + g'g$ instead of $f'g +g'f$. $\endgroup$ Commented May 21 at 15:41
  • $\begingroup$ A pity I didnt notice it, looks so obvious when you see it:)) thanks! $\endgroup$ Commented May 21 at 15:42

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