Efficient/faster methods to find the general closed form of $\int _0^1\frac{\ln \left(ax^2+b\right)}{x+1}\:dx$

Solution 1:

The value of both $a$ and $b$ don't matter as long as $a,b >0$. We can split it up to get that

$$\int_0^1\frac{\log(ax^2+b)}{x+1}\:dx = \log a \log 2 + \int_0^1\frac{\log(x^2+c)}{x+1}\:dx$$

where we have a new parameter $c \equiv \frac{b}{a}$. Then taking the derivative we have that

$$I'(c) = \int_0^1\frac{1}{(x^2+c)(x+1)}\:dx = \frac{1}{c+1}\int_0^1 \frac{1}{x+1}-\frac{x-1}{x^2+c}\:dx$$

$$= \frac{\log 2}{c+1} - \frac{\log c}{2(c+1)} + \frac{\arctan\left(\frac{1}{\sqrt{c}}\right)}{\sqrt{c}(c+1)}$$

which can be solved in similar ways as before, but now it's only one variable.

For slight completeness sake, continuing on a bit further we get that

$$I(a,b) = \log 2 \log(a+b) - \arctan^2\left(\sqrt{\frac{a}{b}}\right)+\int_0^1 \frac{2\log t}{t+1}\:dx - \frac{1}{2}\int_0^{\frac{b}{a}} \frac{\log t}{t+1}\:dt$$

where the value of the last two integrals can be given by special functions.