Using Leibniz's Rule to Evaluate Integrals

Problem Evaluate the integral

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

using Leibniz's rule.

Solution attempt.

To evaluate the above integral, we consider a similar integral

$$F(y) = \int_0^1 \frac{\ln(xy+1)}{x^2+1} \, dx$$

Using Leibniz's rule, we obtain

$$F'(y) = \int_0^1 \frac{\partial f}{\partial y} \Big[ \frac{\ln(xy+1)}{x^2+1} \Big] \, dx = \int_0^1 \frac{1}{x^2+1} \cdot \frac{\partial f}{\partial y} \big[ \ln(xy+1) \big] \, dx$$

$$F'(y) = \int_0^1 \frac{1}{x^2+1} \cdot \frac{x}{xy+1}\, dx = \int_0^1 \frac{x}{(x^2+1)(xy+1)}$$

Where should I go from here? Would it be helpful to let $y=0$ and integrate that somehow? I'm stuck.

Any advice or help is greatly appreciated in advance.

$$I=\int_0^1 \frac{\ln(x+1)}{x^2+1} \, dx$$ $$F(y) = \int_0^1 \frac{\log(xy+1)}{x^2+1} \, dx$$

$$F'(y) = \int_0^1 \frac{x}{(x^2+1)(xy+1)}\, dx=\frac 1 {y^2+1}\int_0^1\Bigg[\frac{y}{x^2+1}+\frac{x}{x^2+1}-\frac{y}{x y+1} \Bigg]\,dx $$ $$\int\Bigg[\frac{y}{x^2+1}+\frac{x}{x^2+1}-\frac{y}{x y+1} \Bigg]\,dx=y \tan ^{-1}(x)+\frac{1}{2} \log \left(x^2+1\right)-\log (x y+1) $$ $$\int_0^1\Bigg[\frac{y}{x^2+1}+\frac{x}{x^2+1}-\frac{y}{x y+1} \Bigg]\,dx=\frac{1}{4} (\pi y-4 \log (y+1)+2\log (2))$$ $$F'(y)=\frac{\pi y}{4 \left(y^2+1\right)}-\frac{\log (y+1)}{y^2+1}+\frac{\log (2)}{2 \left(y^2+1\right)}$$ Now, integrate from $0$ to $1$ $$F(y)=\int_0^1 F'(y)\,dy=\frac \pi 4\int_0^1\frac{ y}{\left(y^2+1\right)}\,dy-I+\frac{\log (2)}{2}\int_0^1 \frac {dy}{y^2+1}\,dy$$ $$F(y)=\frac{1}{8} \pi \log (2)-I+\frac{1}{8} \pi \log (2)\implies F(y)+I=2I=\frac{1}{4} \pi \log (2)$$ $$I=\frac{1}{8} \pi \log (2)$$

Edit

If, as @Jose Jimenez answered, we use $y^2$ instead of $y$, we shall have $$F'(y)=\int_0^1 \frac{2 x y}{\left(x^2+1\right) \left(x y^2+1\right)}\,dx=\frac 2{1+y^4}\int_0^1 \Bigg[\frac{y^3}{x^2+1}+\frac {x y}{x^2+1} -\frac{y^3}{x y^2+1}\Bigg]\,dx$$ $$\int \frac{y^3}{x^2+1}\,dx=y \tan ^{-1}(x)\implies \int_0^1 \frac{y^3}{x^2+1}\,dx=\frac{\pi }{4}y$$ $$\int \frac {x y}{x^2+1}\,dx=\frac{1}{2} y \log \left(x^2+1\right)\implies \int_0^1 \frac {x y}{x^2+1}\,dx=\frac{\log (2)}{2} y $$ $$\int \frac{y^3}{x y^2+1}\,dx=y \log \left(x y^2+1\right)\implies \int_0^1 \frac{y^3}{x y^2+1}\,dx=y \log \left(y^2+1\right)$$ So, now $$F(y)=\int_0^1\Bigg[\frac{\pi y^3}{2 (y^4+1)}+\frac{y \log (2)}{y^4+1}-\frac{2 y \log \left(y^2+1\right)}{y^4+1}\Bigg]\,dy$$

The first and second antiderivatives and integrals do not make any problem $$\int\frac{\pi y^3}{2 (y^4+1)}\,dy=\frac{1}{8} \pi \log \left(y^4+1\right)\implies \int_0^1\frac{\pi y^3}{2 (y^4+1)}\,dy=\frac{1}{8} \pi \log (2)$$ $$\int \frac{y \log (2)}{y^4+1}\,dy=\frac{1}{2} \log (2) \tan ^{-1}\left(y^2\right)\implies \int_0^1\frac{y \log (2)}{y^4+1}\,dy=\frac{1}{8} \pi \log (2)$$ However, the third antiderivative is quite unpleasant $$\int\frac{2 y \log \left(y^2+1\right)}{y^4+1}\,dy$$ involves polylogarithms and complex arguments (even Wolfram Alpha does not give the formal answer). Fortunately $$\int_0^1\frac{2 y \log \left(y^2+1\right)}{y^4+1}\,dy=\frac{1}{8} \pi \log (2)$$ It is amazing to find three identical terms and, by the end, $$I=\frac{1}{8} \pi \log (2)$$

I don't think the way you use Leibniz's rule helps here. I would look at the first example here to help you solve this.