# Finding value of $\int^{2}_{0}\frac{\ln(1+x)}{1+x^2}dx$

Finding value of $\displaystyle \int^{2}_{0}\frac{\ln(1+x)}{1+x^2}dx$

Try: Let $\displaystyle I(a) = \int^{2}_{0}\frac{\ln(1+ax)}{1+x^2}dx$

Then $\displaystyle I'(a) = \int^{2}_{0}\frac{1}{(1+ax)(1+x^2)}dx$

$\displaystyle = \frac{1}{1+a^2}\int^{2}_{0}\bigg[\frac{-a}{1+ax}+\frac{a+x}{1+x^2}\bigg]dx$

$\displaystyle \Rightarrow I'(a) = \frac{1}{1+a^2}\bigg[-\ln(1+2a)+a\tan^{-1}(2)+\frac{\ln(5)}{2}\bigg]$

Could some help me to solve it, Thanks

• I am afraid that this could also involve polylogarithm functions. – Claude Leibovici Sep 20 '18 at 10:58
• Possible duplicate of Evaluate the integral: $\int_{0}^{1} \frac{\ln(x+1)}{x^2+1} \mathrm dx$ – Dinesh Shankar Sep 20 '18 at 12:00
• No it is not a duplicate @Dinesh Shankar. – DXT Sep 20 '18 at 12:12
• Try the substitution $x=\frac {2-w}{1+2w}$ and see if you can make sense of the resulting integral. Maybe you'll notice something I haven't noticed. Another possibility would be to split the integral from zero to one and one to two, then make a substitution $z=x-1$ on the second integral and use partial fractions or something to simplify the rest. – Frank W. Sep 20 '18 at 15:48

Unfortunately, the tricks involving differentiation under the integral sign or odd and even functions will not work since those relied on certain symmetries that arise from having 0 and 1 as the limits of the integral.

You will have to make use of dilogarithms and complex numbers to find the antiderivative, and plug in the limits, unfortunately, to find you desired integral.

The dilogarithm can be defined as follows:

$$Li_2(x)=\int_0^x -\frac{\ln(1-x)}{x}dx$$

As a first step, notice that you can factorise $x^2+1=(x+i)(x-i)$

You can perform partial fraction decomposition to break up $\frac{1}{x^2+1}$:

$$\frac{1}{x^2+1}=\frac{i}{2}(\frac{1}{x+i} - \frac{1}{x-i})$$

So the integral now becomes

$$\int \frac{\ln(x+1)}{x^2+1}dx=\frac{i}{2}\int \frac{\ln(x+1)}{x+i}dx - \frac{i}{2}\int \frac{\ln(x+1)}{x-i}dx$$

I will just solve the first of these. The second can be tackled in a similar fashion, and I will supply the details if needed.

To begin, let us make the substitution $u=x+i, du=dx$ ($u=x-i$ for the other integral), in order to isolate a $u$ in the denominator (to shape the integrand into $\frac{\ln(1-u)}{u}$, so that we can eventually apply the dilogarithm)

$$\int \frac{\ln(x+1)}{x+i}dx = \int \frac{\ln(u+1-i)}{u}du$$

Now we need to somehow transform the expression $\ln(u+1-i)$ in the numerator into the form $\ln(1-w)$

To do this, we remove a factor of $1-i$ from the logarithm, and using the logarithm properties we split the integral up.

$$\int \frac{\ln(u+1-i)}{u}du = \int \frac{\ln(\frac{u}{1-i}+1)}{u}du + \int \frac{\ln(1-i)}{u}du$$

The second integral here is just $\int \frac{\ln(1-i)}{u}du=\ln(1-i)\int \frac{1}{u}du=\ln(1-i)\ln|u|=\ln(1-i)\ln|x+i|$

Now substitute $w=-\frac{u}{1-i}, du=(i-1)dw$ into the other integral.

$$\int \frac{\ln(\frac{u}{1-i}+1)}{u}du=-\int \frac{\ln(1-w)}{w}dw=-Li_2(w)=-Li_2(\frac{x+i}{i-1})$$

Hence

$$\frac{i}{2}\int \frac{\ln(x+1)}{x+i}dx=\frac{i}{2}\ln(1-i)\ln|x+i|-\frac{i}{2}Li_2(\frac{x+i}{i-1})$$

Now after applying a similar tactic with the other integral, we have

$$\int \frac{\ln(x+1)}{x^2+1}dx=\frac{i}{2}\ln(1-i)\ln|x+i|-\frac{i}{2}Li_2(\frac{x+i}{i-1})-\frac{i}{2}\ln(1-i)\ln|x-i|+\frac{i}{2}Li_2(\frac{x-i}{i-1})+C$$

Now just substitute in your limits