WB JEE 2009 | Indefinite Integrals Question 4 | Mathematics

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

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

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

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

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

WB JEE 2009

MCQ (Single Correct Answer)

-0.25

$$\int {{{{{\sin }^{ - 1}}x} \over {\sqrt {1 - {x^2}} }}dx} $$ equal to

where c is an arbitrary constant

$$\log ({\sin ^{ - 1}}x) + c$$

$${1 \over 2}{({\sin ^{ - 1}}x)^2} + c$$

$$\log \left( {\sqrt {1 - {x^2}} } \right) + c$$

$$\sin ({\cos ^{ - 1}}x) + c$$

WB JEE 2010

MCQ (Single Correct Answer)

-0.25

$$\int {{{\log \sqrt x } \over {3x}}dx} $$ is equal to

$${1 \over 3}{\left( {\log \sqrt x } \right)^2} + C$$

$${2 \over 3}{\left( {\log \sqrt x } \right)^2} + C$$

$${2 \over 3}{\left( {\log x} \right)^2} + C$$

$${1 \over 3}{\left( {\log x} \right)^2} + C$$

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