Question

f(x)=ln(x+1)\x^2+1 Subpunctul b) si c)

Answer 1

Explicație pas cu pas:

$\int \left(f(x) + \frac{\arctg(x)}{x + 1}\right) dx = \int \left( \frac{ln(x+1)}{ {x}^{2} + 1} + \frac{\arctg(x)}{x + 1} \right) dx$

$\int fg' = fg - \int f'g$

$f = \arctg(x) => f' = \frac{1}{ {x}^{2} + 1}$

$g' = \frac{1}{x + 1} => g = ln(x + 1)$

→

$\int \left(\frac{\arctg(x)}{x + 1} \right) dx = \int\left(\arctg(x)\cdot \frac{1}{x + 1} \right) dx \\ = \arctg(x)\cdot ln(x + 1) - \int\left( \frac{1}{ {x}^{2} + 1} \cdot ln(x + 1) \right) dx\\ = \arctg(x)\cdot ln(x + 1) - \int\left( \frac{ln(x + 1)}{ {x}^{2} + 1} \right) dx$

⇒

$\int \left(f(x) + \frac{\arctg(x)}{x + 1}\right) dx = \int \left( \frac{ln(x+1)}{ {x}^{2} + 1} \right) dx + \int \left(\frac{\arctg(x)}{x + 1} \right) dx \\ = \int \left( \frac{ln(x+1)}{ {x}^{2} + 1} \right) dx + \arctg(x)\cdot ln(x + 1) - \int\left( \frac{ln(x + 1)}{ {x}^{2} + 1} \right) dx \\ = \arctg(x)\cdot ln(x + 1) + c$

$\int_{0}^{1} \left(f(x) + \frac{\arctg(x)}{x + 1}\right) dx = \arctg(x)\cdot ln(x + 1)|_{0}^{1} \\ = \arctg(1)\cdot ln(1 + 1) - \arctg(0)\cdot ln(0 + 1) \\ = \frac{\pi}{4}\cdot ln(2) - 0 = \frac{\pi ln(2)}{4}$