1
JEE Main 2019 (Online) 10th April Morning Slot
+4
-1
If $$\int {{{dx} \over {{{\left( {{x^2} - 2x + 10} \right)}^2}}}} = A\left( {{{\tan }^{ - 1}}\left( {{{x - 1} \over 3}} \right) + {{f\left( x \right)} \over {{x^2} - 2x + 10}}} \right) + C$$

where C is a constant of integration then :
A
A =$${1 \over {54}}$$ and f(x) = 9(x–1)2
B
A =$${1 \over {54}}$$ and f(x) = 3(x–1)
C
A =$${1 \over {81}}$$ and f(x) = 3(x–1)
D
A =$${1 \over {27}}$$ and f(x) = 9(x–1)2
2
JEE Main 2019 (Online) 9th April Evening Slot
+4
-1
$$\int {{e^{\sec x}}}$$ $$(\sec x\tan xf(x) + \sec x\tan x + se{x^2}x)dx$$
= esecxf(x) + C then a possible choice of f(x) is :-
A
x sec x + tan x + 1/2
B
sec x + xtan x - 1/2
C
sec x - tan x - 1/2
D
sec x + tan x + 1/2
3
JEE Main 2019 (Online) 9th April Morning Slot
+4
-1
The integral $$\int {{\rm{se}}{{\rm{c}}^{{\rm{2/ 3}}}}\,{\rm{x }}\,{\rm{cose}}{{\rm{c}}^{{\rm{4 / 3}}}}{\rm{x \,dx}}}$$ is equal to (Hence C is a constant of integration)
A
-3/4 tan - 4 / 3 x + C
B
3tan–1/3x + C
C
–3cot–1/3x+ C
D
- 3tan–1/3x + C
4
JEE Main 2019 (Online) 8th April Evening Slot
+4
-1
If $$\int {{{dx} \over {{x^3}{{(1 + {x^6})}^{2/3}}}} = xf(x){{(1 + {x^6})}^{{1 \over 3}}} + C}$$
where C is a constant of integration, then the function ƒ(x) is equal to
A
$${3 \over {{x^2}}}$$
B
$$- {1 \over {6{x^3}}}$$
C
$$- {1 \over {2{x^3}}}$$
D
$$- {1 \over {2{x^2}}}$$
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