Explicație pas cu pas:
[tex]\int \left(f(x) + \frac{\arctg(x)}{x + 1}\right) dx = \int \left( \frac{ln(x+1)}{ {x}^{2} + 1} + \frac{\arctg(x)}{x + 1} \right) dx \\ [/tex]
[tex]\int fg' = fg - \int f'g \\ [/tex]
[tex]f = \arctg(x) => f' = \frac{1}{ {x}^{2} + 1} \\ [/tex]
[tex]g' = \frac{1}{x + 1} => g = ln(x + 1) \\ [/tex]
→
[tex]\int \left(\frac{\arctg(x)}{x + 1} \right) dx = \int\left(\arctg(x)\cdot \frac{1}{x + 1} \right) dx \\ = \arctg(x)\cdot ln(x + 1) - \int\left( \frac{1}{ {x}^{2} + 1} \cdot ln(x + 1) \right) dx\\ = \arctg(x)\cdot ln(x + 1) - \int\left( \frac{ln(x + 1)}{ {x}^{2} + 1} \right) dx[/tex]
⇒
[tex]\int \left(f(x) + \frac{\arctg(x)}{x + 1}\right) dx = \int \left( \frac{ln(x+1)}{ {x}^{2} + 1} \right) dx + \int \left(\frac{\arctg(x)}{x + 1} \right) dx \\ = \int \left( \frac{ln(x+1)}{ {x}^{2} + 1} \right) dx + \arctg(x)\cdot ln(x + 1) - \int\left( \frac{ln(x + 1)}{ {x}^{2} + 1} \right) dx \\ = \arctg(x)\cdot ln(x + 1) + c[/tex]
[tex]\int_{0}^{1} \left(f(x) + \frac{\arctg(x)}{x + 1}\right) dx = \arctg(x)\cdot ln(x + 1)|_{0}^{1} \\ = \arctg(1)\cdot ln(1 + 1) - \arctg(0)\cdot ln(0 + 1) \\ = \frac{\pi}{4}\cdot ln(2) - 0 = \frac{\pi ln(2)}{4} [/tex]
