1
JEE Main 2023 (Online) 29th January Evening Shift
+4
-1

The value of the integral $$\int_1^2 {\left( {{{{t^4} + 1} \over {{t^6} + 1}}} \right)dt}$$ is

A
$${\tan ^{ - 1}}{1 \over 2} - {1 \over 3}{\tan ^{ - 1}}8 + {\pi \over 3}$$
B
$${\tan ^{ - 1}}2 - {1 \over 3}{\tan ^{ - 1}}8 + {\pi \over 3}$$
C
$${\tan ^{ - 1}}2 + {1 \over 3}{\tan ^{ - 1}}8 - {\pi \over 3}$$
D
$${\tan ^{ - 1}}{1 \over 2} + {1 \over 3}{\tan ^{ - 1}}8 - {\pi \over 3}$$
2
JEE Main 2023 (Online) 29th January Evening Shift
+4
-1

The value of the integral $$\int\limits_{1/2}^2 {{{{{\tan }^{ - 1}}x} \over x}dx}$$ is equal to :

A
$${\pi \over 2}{\log _e}2$$
B
$${\pi \over 4}{\log _e}2$$
C
$${1 \over 2}{\log _e}2$$
D
$$\pi {\log _e}2$$
3
JEE Main 2023 (Online) 29th January Morning Shift
+4
-1

Let $$f(x) = x + {a \over {{\pi ^2} - 4}}\sin x + {b \over {{\pi ^2} - 4}}\cos x,x \in R$$ be a function which

satisfies $$f(x) = x + \int\limits_0^{\pi /2} {\sin (x + y)f(y)dy}$$. then $$(a+b)$$ is equal to

A
$$- 2\pi (\pi + 2)$$
B
$$- \pi (\pi - 2)$$
C
$$- \pi (\pi + 2)$$
D
$$- 2\pi (\pi - 2)$$
4
JEE Main 2023 (Online) 25th January Evening Shift
+4
-1

The integral $$16\int\limits_1^2 {{{dx} \over {{x^3}{{\left( {{x^2} + 2} \right)}^2}}}}$$ is equal to

A
$${{11} \over {12}} + {\log _e}4$$
B
$${{11} \over 6} + {\log _e}4$$
C
$${{11} \over {12}} - {\log _e}4$$
D
$${{11} \over 6} - {\log _e}4$$
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