int_{-1}^1 frac{sqrt{1+x+x^2}-sqrt{1-x+x^2}}{ | vedclass

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(C) Let $I = \int_{-1}^1 f(x) dx$,where $f(x) = \frac{\sqrt{1+x+x^2}-\sqrt{1-x+x^2}}{\sqrt{1+x+x^2}+\sqrt{1-x+x^2}}$.We check if the function is even

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(C) Let $I = \int_{-1}^1 f(x) dx$,where $f(x) = \frac{\sqrt{1+x+x^2}-\sqrt{1-x+x^2}}{\sqrt{1+x+x^2}+\sqrt{1-x+x^2}}$.We check if the function is even or odd by evaluating $f(-x)$:$f(-x) = \frac{\sqrt{1+(-x)+(-x)^2}-\sqrt{1-(-x)+(-x)^2}}{\sqrt{1+(-x)+(-x)^2}+\sqrt{1-(-x)+(-x)^2}}$$f(-x) = \frac{\sqrt{1-x+x^2}-\sqrt{1+x+x^2}}{\sqrt{1-x+x^2}+\sqrt{1+x+x^2}}$$f(-x) = -\left( \frac{\sqrt{1+x+x^2}-\sqrt{1-x+x^2}}{\sqrt{1+x+x^2}+\sqrt{1-x+x^2}} \right) = -f(x)$.Since $f(-x) = -f(x)$,the function $f(x)$ is an odd function.By the property of definite integrals,if $f(x)$ is an odd function,then $\int_{-a}^a f(x) dx = 0$.Therefore,$\int_{-1}^1 f(x) dx = 0$.

$\int_{-1}^1 \frac{\sqrt{1+x+x^2}-\sqrt{1-x+x^2}}{\sqrt{1+x+x^2}+\sqrt{1-x+x^2}} dx$ is equal to

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