1
BITSAT 2023
+3
-1

Let $$f(x)=\int \frac{\sqrt{x}}{(1+x)^2} d x$$, where $$x \geq 0$$. Then, $$f(3)-f(1)$$ is equal to

A
$$\frac{\pi}{12}+\frac{1}{2}-\frac{\sqrt{3}}{4}$$
B
$$-\frac{\pi}{6}+\frac{1}{2}+\frac{\sqrt{3}}{4}$$
C
$$-\frac{\pi}{12}+\frac{1}{2}+\frac{\sqrt{3}}{4}$$
D
$$\frac{\pi}{6}+\frac{1}{2}-\frac{\sqrt{3}}{4}$$
2
BITSAT 2022
+3
-1

The value of $$\int {{1 \over {{{[{{(x - 1)}^3}{{(x + 2)}^5}]}^{{1 \over 4}}}}}dx}$$, is

A
$${4 \over 3}{\left( {{{x + 1} \over {x - 2}}} \right)^{{1 \over 4}}} + C$$
B
$${3 \over 4}{\left( {{{x - 1} \over {x + 2}}} \right)^{{1 \over 4}}} + C$$
C
$${4 \over 3}{\left( {{{x - 1} \over {x + 2}}} \right)^{{1 \over 4}}} + C$$
D
$${1 \over 3}{\left( {{{2x - 1} \over {4x - 3}}} \right)^{{1 \over 4}}} + C$$
3
BITSAT 2022
+3
-1

Let $$f(x) = \int {{{{x^2}dx} \over {(1 + {x^2})(1 + \sqrt {1 + {x^2}} )}}}$$ and $$f(0) = 0$$, then the value of $$f(1)$$ be

A
$$\log (1 + \sqrt 2 )$$
B
$$\log (1 + \sqrt 2 ) - {\pi \over 4}$$
C
$$\log (1 + \sqrt 2 ) + {\pi \over 2}$$
D
None of these
4
BITSAT 2021
+3
-1

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

A
$${1 \over {2\sqrt 3 }}\log \left| {{{\tan {x \over 2} - 2 - \sqrt 3 } \over {\tan {x \over 2} - 2 + \sqrt 3 }}} \right| + c$$
B
$${{\sqrt 3 } \over 2}\log \left| {{{\tan {x \over 2} - 2 - \sqrt 3 } \over {\tan {x \over 2} - 2 + \sqrt 3 }}} \right| + c$$
C
$${1 \over {\sqrt 3 }}\log \left| {{{\tan {x \over 2} - 2 - \sqrt 3 } \over {\tan {x \over 2} - 2 + \sqrt 3 }}} \right| + c$$
D
None of the above
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