1
JEE Main 2019 (Online) 12th April Morning Slot
+4
-1
The integral $$\int {{{2{x^3} - 1} \over {{x^4} + x}}} dx$$ is equal to :
(Here C is a constant of integration)
A
$${\log _e}{{\left| {{x^3} + 1} \right|} \over {{x^2}}} + C$$
B
$${1 \over 2}{\log _e}{{\left| {{x^3} + 1} \right|} \over {{x^2}}} + C$$
C
$${\log _e}\left| {{{{x^3} + 1} \over x}} \right| + C$$
D
$${1 \over 2}{\log _e}{{{{\left( {{x^3} + 1} \right)}^2}} \over {\left| {{x^3}} \right|}} + C$$
2
JEE Main 2019 (Online) 10th April Evening Slot
+4
-1
If $$\int {{x^5}} {e^{ - {x^2}}}dx = g\left( x \right){e^{ - {x^2}}} + c$$, where c is a constant of integration, then $$g$$(–1) is equal to :
A
1
B
- 1
C
$$- {5 \over 2}$$
D
$$- {1 \over 2}$$
3
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
4
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
EXAM MAP
Medical
NEET
GATE CSEGATE ECEGATE EEGATE MEGATE CEGATE PIGATE IN
Civil Services
UPSC Civil Service
Defence
NDA
CBSE
Class 12