The value of the integral is equal to: — JEE PYQs | Mockzy

HardJEE PYQs

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The value of the integral $\int_{0}^{2} \frac{x ( x ^{2} + x + 1 )}{x + 1 x ^{4} + x ^{2} + 1} d x$ is equal to:

$\frac{1}{3} lo g_{e} (3 - 22)$ .

$\frac{2}{3} lo g_{e} (4 + 2)$ .

$\frac{2}{3} lo g_{e} (3 + 22)$ .

$\frac{1}{3} lo g_{e} (1 + 62)$ .

Let the foot of the perpendicular from the point (λ,2,3) on the line be the poin

Let y = y(x) be the solution of the differential equation +, where ,, . Then is

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