WB JEE 2009 | Indefinite Integrals Question 2 | Mathematics

WB JEE 2009

MCQ (Single Correct Answer)

-0.25

$$\int {{{dx} \over {x(x + 1)}}} $$ equals

where c is arbitrary constant.

$$\ln \left| {{{x + 1} \over x}} \right| + c$$

$$\ln \left| {{x \over {x + 1}}} \right| + c$$

$$\ln \left| {{{x - 1} \over x}} \right| + c$$

$$\ln \left| {{{x - 1} \over {x + 1}}} \right| + c$$

WB JEE 2009

MCQ (Single Correct Answer)

-0.25

The value of $$\mathop {\lim }\limits_{n \to \infty } \left[ {{n \over {{n^2} + {1^2}}} + {n \over {{n^2} + {2^2}}} + ..... + {n \over {{n^2} + {n^2}}}} \right]$$ is

$$\pi$$ / 4

log 2

WB JEE 2009

MCQ (Single Correct Answer)

-0.25

The value of $$\int\limits_{ - 1}^1 {{{|x + 2|} \over {x + 2}}dx} $$ is

$$-$$1

WB JEE 2009

MCQ (Single Correct Answer)

-0.25

$$\int {{{dx} \over {\sin x + \sqrt 3 \cos x}}} $$ equals

$${1 \over 2}\ln \left| {\tan \left( {{x \over 2} - {\pi \over 6}} \right)} \right| + c$$

$${1 \over 2}\ln \left| {\tan \left( {{x \over 4} - {\pi \over 6}} \right)} \right| + c$$

$${1 \over 2}\ln \left| {\tan \left( {{x \over 2} + {\pi \over 6}} \right)} \right| + c$$

$${1 \over 2}\ln \left| {\tan \left( {{x \over 4} + {\pi \over 6}} \right)} \right| + c$$

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