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$∫\frac{x^4-1}{x^2(x^4+x^2+1)^{1/2}}dx$ is equal to |
$\sqrt{\frac{x^4+x^2+1}{x}}+c$ $\sqrt{x^2+1\frac{1}{x^2}}+c$ $\sqrt{\frac{x^4+x^2+1}{x^2}}+c$ none of these |
$\sqrt{\frac{x^4+x^2+1}{x^2}}+c$ |
Let $I=∫\frac{x^4-1}{x^2(x^4+x^2+1)^{1/2}}dx=∫\frac{x^3(x-\frac{1}{x^3})dx}{x^3\sqrt{x^2+\frac{1}{x^2}+1}}$ If $x^2+\frac{1}{x^2}+1=p^2$, then $(2x-\frac{2}{x^3})dx=2pdp$ $⇒I=∫\frac{pdp}{p}p+c=\sqrt{\frac{x^4+x^2+1}{x^2}}+c$ |
