# Computing an integral using differential under the integral sign

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!!

• 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}
– NEON
Commented May 21 at 15:28
• @NEON it was a question on our vector analysis exam, and I got stuck at this part :) Commented May 21 at 15:42

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?
• I take it you were looking for something of the form $f'f + g'g$ instead of $f'g +g'f$. Commented May 21 at 15:41