[tex]f(x) = \dfrac{x^3+5x}{x^2+1} \\ \\ \displaystyle I = \int_{0}^3 f^{-1}(t)\, dt \\ \\ \\t = \dfrac{x^3+5x}{x^2+1} \Rightarrow t = \dfrac{x(x^2+1)+4x}{x^2+1} = x+\dfrac{4x}{x^2+1}\Rightarrow \\ \\ \Rightarrow dt = \Bigg(1 + \dfrac{4(x^2+1)-4x\cdot 2x}{(x^2+1)^2}\Bigg) \, dx \Rightarrow dt = \Bigg(1+ \dfrac{-4x^2+4}{(x^2+1)^2} \Bigg)\, dx \Rightarrow \\ \\ \Rightarrow dt = \Bigg(1-\dfrac{4(x-1)(x+1)}{(x^2+1)^2}\Bigg)\, dx\\ \\ t = 0\Rightarrow x = 0\\ t = 3\Rightarrow x = 1\\ \\ \text{Se gasesc usor valorile, fara a trebui calculate.}[/tex]
[tex]\displaystyle \Rightarrow I = \int_{0}^1 f^{-1}\Big(f(x)\Big)\cdot \Big( 1-\dfrac{4(x-1)(x+1)}{(x^2+1)^2}\Big)\, dx = \\ \\ = \int_{0}^1 x\cdot \Big( 1-\dfrac{4(x-1)(x+1)}{(x^2+1)^2}\Big)\, dx\\ \\ \\x^2+1 = y \Rightarrow 2x \, dx = dy \\ x^2 = y-1 \Rightarrow x = \sqrt{y-1} \\ x = 0 \Rightarrow y = 1 \\ x = 1 \Rightarrow y = 2[/tex]
[tex]\displaystyle I = \dfrac{1}{2}\int_{1}^2\Big(1 - \dfrac{4(y-1-1)}{y^2}\Big)\, dy = \dfrac{1}{2}\Big(y\Big|_{1}^2 - 4\ln y \Big|_{1}^2+\dfrac{8y^{-1}}{-1}\Big) = \\ \\ = \dfrac{1}{2}\Big(1-4\ln 2 -4+8\Big) = \dfrac{1}{2}\Big(5-4\ln 2\Big)[/tex]
=> a) corect
