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Recently I've tackled the Putnam 2005 A5 integral, that is, $\displaystyle\int_{0}^{1}\frac{\ln(x+1)}{x^2+1}\mathrm{d}x$. In solving this, I converted it to multiple integrals- I used a $u$ substitution to show it was equal to $\displaystyle\int_{0}^{\frac{\pi}{4}}\ln(\sin{x}+\cos{x})-\ln(\cos{x})~\mathrm{d}x$. This enabled me to make a symmetry argument and it allowed me to finally solve it.

This made me wonder, does converting an integral into multiple always make it easier to solve? This is the central idea of partial fractions. This integral also finally simplified once I converted it into multiple integrals. Do pretty much all integrals become easier, or at least not harder, after making them into multiple integrals?

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    $\begingroup$ $\int f(x)dx=\int f(x)+e^{x^2}-e^{x^2}dx$ $\endgroup$ – vadim123 Feb 4 '18 at 3:29
  • $\begingroup$ @vadim123 Easy, that's $F(x)$ $\endgroup$ – Sudix Feb 4 '18 at 3:43
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In general, no - a sum of integrals is as "hard" to solve as its hardest term, so you should look for ways to simplify the most difficult term, whether that be by combining it with others, splitting it apart or something else.

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A conversion $\int\mapsto\iint $ can be performed in many ways, so your question is actually pretty vague. On the other hand Feynman's trick can be regarded as a standard way for performing such a conversion and it tends to work pretty good, especially if the original integrand function involves a logarithm/arctangent.

In our case $ \int_{0}^{\pi/2}\log(1+\tan\theta)\,d\theta=\frac{\pi}{8}\log 2$ can be computed in a simple way from the Fourier series of $\log\sin$ and $\log\cos$, too.

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No need to use Feynman's trick (or multiple integral) to compute:

$\displaystyle J=\int_0^1 \frac{\ln(1+x)}{1+x^2}\,dx$

Perform the change of variable $y=\dfrac{1-x}{1+x}$,

$\begin{align}J&=\int_0^1 \frac{\ln\left(\frac{2}{1+y}\right)}{1+y^2}\,dy\\ &=\ln 2\int_0^1 \frac{1}{1+y^2}\,dy-J\\ &=\ln 2\Big[\arctan y\Big]_0^1-J\\ &=\frac{\pi}{4}\ln 2 -J \end{align}$

Therefore,

$\boxed{\displaystyle J=\frac{\pi}{8}\ln 2}$

BUT, i don't know a way to compute:

$\displaystyle \int_0^1 \frac{\arctan x}{x\sqrt{1-x^2}}\,dx$

Using only change of variable $y=\phi(x)$ and integration by parts.

(see Definite integral $\int_0^1 \frac{\arctan x}{x\,\sqrt{1-x^2}}\,\text{d}x$ )

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