WB JEE 2009 | Indefinite Integrals Question 6 | Mathematics

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

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

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

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

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

WB JEE 2010

MCQ (Single Correct Answer)

-0.25

$$\int {{e^x}\left( {{2 \over x} - {2 \over {{x^2}}}} \right)dx} $$ is equal to

$${{{e^x}} \over x} + C$$

$${{{e^x}} \over {2{x^2}}} + C$$

$${{2{e^x}} \over x} + C$$

$${{2{e^x}} \over {{x^2}}} + C$$

WB JEE 2010

MCQ (Single Correct Answer)

-0.25

The value of the integral $$\int {{{dx} \over {{{({e^x} + {e^{ - x}})}^2}}}} $$ is

$${1 \over 2}({e^{2x}} + 1) + C$$

$${1 \over 2}({e^{ - 2x}} + 1) + C$$

$$ - {1 \over 2}{({e^{2x}} + 1)^{ - 1}} + C$$

$${1 \over 4}({e^{2x}} - 1) + C$$

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