int frac{x^4-1}{x^2 sqrt{x^4+x^2+1}} , dx = | vedclass

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Question

(A) We have $I = \int \frac{x^4-1}{x^2 \sqrt{x^4+x^2+1}} \, dx$. Divide the numerator and denominator inside the square root by $x^2$: $I = \int \frac

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$\frac{\sqrt{x^4+x^2+1}}{x} + c$

Vedclass · Answer

(A) We have $I = \int \frac{x^4-1}{x^2 \sqrt{x^4+x^2+1}} \, dx$. Divide the numerator and denominator inside the square root by $x^2$: $I = \int \frac{x^2 - \frac{1}{x^2}}{\sqrt{x^2 + 1 + \frac{1}{x^2}}} \, dx$. Let $t = x + \frac{1}{x}$. Then $dt = (1 - \frac{1}{x^2}) \, dx$. Note that $t^2 = (x + \frac{1}{x})^2 = x^2 + 2 + \frac{1}{x^2}$,so $x^2 + \frac{1}{x^2} = t^2 - 2$. Substituting these into the integral: $I = \int \frac{dt}{\sqrt{t^2 - 2 + 1}} = \int \frac{dt}{\sqrt{t^2 - 1}}$. This is a standard integral: $\int \frac{dt}{\sqrt{t^2 - 1}} = \ln|t + \sqrt{t^2 - 1}| + c$. However,looking at the options,let's re-evaluate the derivative of the expression $\frac{\sqrt{x^4+x^2+1}}{x}$. Let $f(x) = \frac{\sqrt{x^4+x^2+1}}{x} = \sqrt{x^2 + 1 + \frac{1}{x^2}}$. $f'(x) = \frac{1}{2\sqrt{x^2 + 1 + \frac{1}{x^2}}} \cdot (2x - \frac{2}{x^3}) = \frac{x - \frac{1}{x^3}}{\sqrt{x^2 + 1 + \frac{1}{x^2}}} = \frac{\frac{x^4-1}{x^3}}{\frac{\sqrt{x^4+x^2+1}}{x}} = \frac{x^4-1}{x^2 \sqrt{x^4+x^2+1}}$. Thus,the integral is $\frac{\sqrt{x^4+x^2+1}}{x} + c$.

$\int \frac{x^4-1}{x^2 \sqrt{x^4+x^2+1}} \, dx =$

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