CUET Preparation Today
$\int\left(\sqrt{x}-\frac{1}{\sqrt{x}}\right)^2 d x$ is equal to |
$\frac{x^2}{2}+\log |x|+c$ $\frac{x^2}{2}+\log |x|+2 x+c$ $\frac{1}{3}\left(\sqrt{x}+\frac{1}{\sqrt{x}}\right)^3+c$ $\frac{x^2}{2}+\log x-2 x+c$ |
$\frac{x^2}{2}+\log x-2 x+c$ |
$\int\left(\sqrt{x}-\frac{1}{\sqrt{x}}\right)^2 d x=\int\left(x+\frac{1}{x}-2\right) d x=\frac{x^2}{2}+\ln |x|-2 x+c$ Hence (4) is the correct answer. |
