Wiskunde Multiple Choice Vraag uit Gricha's Wiskundige Vragenbank

$\int_{-1}^{+1} \frac{1}{(1+x^2)(e^x+1)}\; dx=$	A. 1
B. π
C.
D.
E.

$I=\int_{-1}^{+1}\frac{1}{(1+x^2)(e^x+1)}\: dx \begin{matrix} {\color{Yellow} .}_{y=-x} \\ \begin{matrix} =\\ {\color{Yellow} .} \end{matrix} \end{matrix} \int_{+1}^{-1}\frac{1}{(1+y^2)(e^{-y}+1)}\: d(\!-y) \newline {\color{Yellow} .}\quad=\int_{-1}^{+1}\frac{1}{(1+y^2)e^{-y}(1+e^y)}\: dy =\int_{-1}^{+1}\frac{e^x}{(1+x^2)(1+e^x)}\: dx\ \newline \Rightarrow I+I=\int_{-1}^{+1}\frac{1+e^x}{(1+x^2)(1+e^x)}\: dx =\int_{-1}^{+1}\frac{1}{1+x^2}\: dx \newline {\color{Yellow} .}\qquad \quad=\left [ Bgtan\: x \right ]_{-1}^{+1}=\frac{\pi }{4}-(\frac{\pi }{4})=\frac{\pi }{2} \newline \Rightarrow I=\frac{\pi }{4} \quad \approx 0,7854$