$∫\frac{(x^2-1)}{(x^2+1)\sqrt{x^4+1}}dx$ is equal to |
$sec^{-1}(\frac{x^2+1}{\sqrt{2}x})+c$ $\frac{1}{\sqrt{2}}sec^{-1}(\frac{x^2+1}{\sqrt{2}x})+c$ $\frac{1}{\sqrt{2}}sec^{-1}(\frac{x^2+1}{\sqrt{2}})+c$ none of these. |
$\frac{1}{\sqrt{2}}sec^{-1}(\frac{x^2+1}{\sqrt{2}x})+c$ |
Let $I=∫\frac{x^2(1-\frac{1}{x^2})dx}{x^2(x+\frac{1}{x})(x^2+\frac{1}{x^2})^{1/2}}$. Let $x+\frac{1}{x}=p⇒(1-\frac{1}{x^2})dx=dp$ $⇒I=∫\frac{dp}{p\sqrt{p^2-2}}=\frac{1}{\sqrt{2}}sec^{-1}\frac{p}{\sqrt{2}}+c=\frac{1}{\sqrt{2}}sec^{-1}(\frac{x^2+1}{\sqrt{2}x})+c$ Hence (B) is the correct answer. |