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JEE Main Previous Year Questions (2026): Indefinite Integral
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Page 1
JEE Main Previous Year Questions
(2025): Indefinite Integral
Q1: If ? ?? ?? (
?? ?? ?? ?? - ?? ? ?? v
?? - ?? ?? +
?? ?? ?? - ?? ? ?? ( ?? - ?? ?? )
?? / ?? +
?? ?? - ?? ?? ) ?? ?? = ?? ( ?? ) + ?? , where C is the constant of
integration, then ?? (
?? ?? ) equals :
JEE Main 2025 (Online) 8th April Evening Shift
Options:
A. 2 + 3 ??
B. 4 ? + ??
C. 1 + 3 ??
D. 3 + 2 ??
Ans: C
Solution:
Let, I = ?? 2
?
- 1
3 / 2
? | x s i n ? ?? x | dx
= ?? 2
{ ?
- 1
1
? x s i n ? ?? x dx - ?
1
3 / 2
? x s i n ? ?? x dx }
= ?? 2
{ 2 ?
0
1
? x s i n ? ?? x dx - ?
- 1
3 / 2
? x s i n ? ?? x dx }
Consider
? x s i n ? ?? x dx
- x ·
1
?? c os ? ?? x + ? 1 ·
1
?? c os ? ?? x dx
= - -
x
?? c os ? ?? x +
sin ? ?? x
?? 2
I = ?? 2
{ 2 ( -
x
?? c os ? ?? x +
sin ? ?? x
?? 2
)
0
1
- ( -
x
?? c os ? ?? x +
sin ? ?? x
?? 2
)
1
3 / 2
}
= ?? 2
{
2
?? - ( -
1
?? 2
-
1
?? ) }
= ?? 2
{
3
?? +
1
?? 2
}
= 3 ?? + 1
Page 2
JEE Main Previous Year Questions
(2025): Indefinite Integral
Q1: If ? ?? ?? (
?? ?? ?? ?? - ?? ? ?? v
?? - ?? ?? +
?? ?? ?? - ?? ? ?? ( ?? - ?? ?? )
?? / ?? +
?? ?? - ?? ?? ) ?? ?? = ?? ( ?? ) + ?? , where C is the constant of
integration, then ?? (
?? ?? ) equals :
JEE Main 2025 (Online) 8th April Evening Shift
Options:
A. 2 + 3 ??
B. 4 ? + ??
C. 1 + 3 ??
D. 3 + 2 ??
Ans: C
Solution:
Let, I = ?? 2
?
- 1
3 / 2
? | x s i n ? ?? x | dx
= ?? 2
{ ?
- 1
1
? x s i n ? ?? x dx - ?
1
3 / 2
? x s i n ? ?? x dx }
= ?? 2
{ 2 ?
0
1
? x s i n ? ?? x dx - ?
- 1
3 / 2
? x s i n ? ?? x dx }
Consider
? x s i n ? ?? x dx
- x ·
1
?? c os ? ?? x + ? 1 ·
1
?? c os ? ?? x dx
= - -
x
?? c os ? ?? x +
sin ? ?? x
?? 2
I = ?? 2
{ 2 ( -
x
?? c os ? ?? x +
sin ? ?? x
?? 2
)
0
1
- ( -
x
?? c os ? ?? x +
sin ? ?? x
?? 2
)
1
3 / 2
}
= ?? 2
{
2
?? - ( -
1
?? 2
-
1
?? ) }
= ?? 2
{
3
?? +
1
?? 2
}
= 3 ?? + 1
Q2: If ? ?? ?? (
?? ?? ???? - ?? ? ?? v
?? - ?? ?? +
?? ???? - ?? ? ?? ( ?? - ?? ?? )
?? / ?? +
?? ?? - ?? ?? ) ?? ?? = ?? ( ?? ) + ?? , where C is the constant of integration,
then ?? (
?? ?? ) equals :
Options:
A.
?? 6
v
e
3
B.
?? 6
v
e
2
C.
?? 4
v
e
3
D.
?? 4
v
e
2
Ans: A
Solution:
?
d
dx
(
x sin
- 1
? x
v 1 - x
2
) =
s i n
- 1
? x
( 1 - x
2
)
3 / 2
+
x
1 - x
2
? ? e
x
(
x sin
- 1
? x
v 1 - x
2
+
sin
- 1
? x
( 1 - x
2
)
3 / 2
+
x
1 - x
2
) dx
= e
x
·
x sin
- 1
? x
v 1 - x
2
+ c = g ( x ) + C
Note : assuming ?? ( ?? ) =
xe
x
s i n
- 1
? ?? v 1 - x
2
?? ( 1 / 2 ) =
e
1 / 2
2
·
?? 6
× 2
v 3
=
?? 6
v
e
3
Comment : In this Q we will not get a unique function g ( x ) , but in order to match the answer we
will have to assume ?? ( ?? ) =
xe s i n
- 1
? x
v 1 - x
2
.
Q3: Let ?? ( ?? ) = ?
????
( ?? - ???? )
????
???? ( ?? + ???? )
????
????
. If ?? ( ???? ) - ?? ( ???? ) =
?? ?? (
?? ?? ?? ????
-
?? ?? ?? ????
) , ?? , ?? ? ?? , then
?? ( ?? + ?? ) is equal to
JEE Main 2025 (Online) 23rd January Morning Shift
Options:
A. 39
B. 22
Page 3
JEE Main Previous Year Questions
(2025): Indefinite Integral
Q1: If ? ?? ?? (
?? ?? ?? ?? - ?? ? ?? v
?? - ?? ?? +
?? ?? ?? - ?? ? ?? ( ?? - ?? ?? )
?? / ?? +
?? ?? - ?? ?? ) ?? ?? = ?? ( ?? ) + ?? , where C is the constant of
integration, then ?? (
?? ?? ) equals :
JEE Main 2025 (Online) 8th April Evening Shift
Options:
A. 2 + 3 ??
B. 4 ? + ??
C. 1 + 3 ??
D. 3 + 2 ??
Ans: C
Solution:
Let, I = ?? 2
?
- 1
3 / 2
? | x s i n ? ?? x | dx
= ?? 2
{ ?
- 1
1
? x s i n ? ?? x dx - ?
1
3 / 2
? x s i n ? ?? x dx }
= ?? 2
{ 2 ?
0
1
? x s i n ? ?? x dx - ?
- 1
3 / 2
? x s i n ? ?? x dx }
Consider
? x s i n ? ?? x dx
- x ·
1
?? c os ? ?? x + ? 1 ·
1
?? c os ? ?? x dx
= - -
x
?? c os ? ?? x +
sin ? ?? x
?? 2
I = ?? 2
{ 2 ( -
x
?? c os ? ?? x +
sin ? ?? x
?? 2
)
0
1
- ( -
x
?? c os ? ?? x +
sin ? ?? x
?? 2
)
1
3 / 2
}
= ?? 2
{
2
?? - ( -
1
?? 2
-
1
?? ) }
= ?? 2
{
3
?? +
1
?? 2
}
= 3 ?? + 1
Q2: If ? ?? ?? (
?? ?? ???? - ?? ? ?? v
?? - ?? ?? +
?? ???? - ?? ? ?? ( ?? - ?? ?? )
?? / ?? +
?? ?? - ?? ?? ) ?? ?? = ?? ( ?? ) + ?? , where C is the constant of integration,
then ?? (
?? ?? ) equals :
Options:
A.
?? 6
v
e
3
B.
?? 6
v
e
2
C.
?? 4
v
e
3
D.
?? 4
v
e
2
Ans: A
Solution:
?
d
dx
(
x sin
- 1
? x
v 1 - x
2
) =
s i n
- 1
? x
( 1 - x
2
)
3 / 2
+
x
1 - x
2
? ? e
x
(
x sin
- 1
? x
v 1 - x
2
+
sin
- 1
? x
( 1 - x
2
)
3 / 2
+
x
1 - x
2
) dx
= e
x
·
x sin
- 1
? x
v 1 - x
2
+ c = g ( x ) + C
Note : assuming ?? ( ?? ) =
xe
x
s i n
- 1
? ?? v 1 - x
2
?? ( 1 / 2 ) =
e
1 / 2
2
·
?? 6
× 2
v 3
=
?? 6
v
e
3
Comment : In this Q we will not get a unique function g ( x ) , but in order to match the answer we
will have to assume ?? ( ?? ) =
xe s i n
- 1
? x
v 1 - x
2
.
Q3: Let ?? ( ?? ) = ?
????
( ?? - ???? )
????
???? ( ?? + ???? )
????
????
. If ?? ( ???? ) - ?? ( ???? ) =
?? ?? (
?? ?? ?? ????
-
?? ?? ?? ????
) , ?? , ?? ? ?? , then
?? ( ?? + ?? ) is equal to
JEE Main 2025 (Online) 23rd January Morning Shift
Options:
A. 39
B. 22
C. 40
D. 26
Ans: A
Solution:
?? ( ?? ) = ?
????
( ?? - 11 )
11
13
( ?? + 15 )
15
13
Put
?? - 11
?? + 15
= ?? ?
26
( ?? + 5 )
2
???? = ????
?? ( ?? ) =
1
26
?
????
?? 11 / 13
=
1
26
·
?? 2 / 13
2 / 13
I ( x ) =
1
4
(
x - 11
x + 15
)
2 / 13
+ C
I ( 37 ) - I ( 24 ) =
1
4
(
26
52
)
2 /13
-
1
4
(
13
39
)
2 / 13
=
1
4
(
1
2
2 / 13
-
1
3
2 / 13
)
=
1
4
(
1
4
1 / 13
-
1
9
1 /13
)
? b = 4 , c = 9
3 ( b + c ) = 39
Q4: Let ? ?? ?? ?? ???? ? ?? ?? ?? = ?? ( ?? ) + ?? , where ?? is the constant of integration. If
?? ( ?? (
?? ?? ) + ?? '
(
?? ?? ) ) = ?? ?? ?? + ?? ?? ?? + ?? , ?? , ?? , ?? ? ?? , then ?? + ?? - ?? equals :
JEE Main 2025 (Online) 23rd January Evening Shift
Options:
A. 47
B. 55
C. 62
D. 48
Ans: B
Solution:
? ?? 3
sin ? ???? ?? = - ?? 3
c o s ? ?? + ? 3 ?? 2
c o s ? ???? ??
= - ?? 3
c os ? ?? + 3 ?? 2
s i n ? ?? - ? 6 ?? s i n ? ???? ??
= - ?? 3
c os ? ?? + 3 ?? 2
s i n ? ?? + 6 ?? c os ? ?? - 6s i n ? ?? + ??
Page 4
JEE Main Previous Year Questions
(2025): Indefinite Integral
Q1: If ? ?? ?? (
?? ?? ?? ?? - ?? ? ?? v
?? - ?? ?? +
?? ?? ?? - ?? ? ?? ( ?? - ?? ?? )
?? / ?? +
?? ?? - ?? ?? ) ?? ?? = ?? ( ?? ) + ?? , where C is the constant of
integration, then ?? (
?? ?? ) equals :
JEE Main 2025 (Online) 8th April Evening Shift
Options:
A. 2 + 3 ??
B. 4 ? + ??
C. 1 + 3 ??
D. 3 + 2 ??
Ans: C
Solution:
Let, I = ?? 2
?
- 1
3 / 2
? | x s i n ? ?? x | dx
= ?? 2
{ ?
- 1
1
? x s i n ? ?? x dx - ?
1
3 / 2
? x s i n ? ?? x dx }
= ?? 2
{ 2 ?
0
1
? x s i n ? ?? x dx - ?
- 1
3 / 2
? x s i n ? ?? x dx }
Consider
? x s i n ? ?? x dx
- x ·
1
?? c os ? ?? x + ? 1 ·
1
?? c os ? ?? x dx
= - -
x
?? c os ? ?? x +
sin ? ?? x
?? 2
I = ?? 2
{ 2 ( -
x
?? c os ? ?? x +
sin ? ?? x
?? 2
)
0
1
- ( -
x
?? c os ? ?? x +
sin ? ?? x
?? 2
)
1
3 / 2
}
= ?? 2
{
2
?? - ( -
1
?? 2
-
1
?? ) }
= ?? 2
{
3
?? +
1
?? 2
}
= 3 ?? + 1
Q2: If ? ?? ?? (
?? ?? ???? - ?? ? ?? v
?? - ?? ?? +
?? ???? - ?? ? ?? ( ?? - ?? ?? )
?? / ?? +
?? ?? - ?? ?? ) ?? ?? = ?? ( ?? ) + ?? , where C is the constant of integration,
then ?? (
?? ?? ) equals :
Options:
A.
?? 6
v
e
3
B.
?? 6
v
e
2
C.
?? 4
v
e
3
D.
?? 4
v
e
2
Ans: A
Solution:
?
d
dx
(
x sin
- 1
? x
v 1 - x
2
) =
s i n
- 1
? x
( 1 - x
2
)
3 / 2
+
x
1 - x
2
? ? e
x
(
x sin
- 1
? x
v 1 - x
2
+
sin
- 1
? x
( 1 - x
2
)
3 / 2
+
x
1 - x
2
) dx
= e
x
·
x sin
- 1
? x
v 1 - x
2
+ c = g ( x ) + C
Note : assuming ?? ( ?? ) =
xe
x
s i n
- 1
? ?? v 1 - x
2
?? ( 1 / 2 ) =
e
1 / 2
2
·
?? 6
× 2
v 3
=
?? 6
v
e
3
Comment : In this Q we will not get a unique function g ( x ) , but in order to match the answer we
will have to assume ?? ( ?? ) =
xe s i n
- 1
? x
v 1 - x
2
.
Q3: Let ?? ( ?? ) = ?
????
( ?? - ???? )
????
???? ( ?? + ???? )
????
????
. If ?? ( ???? ) - ?? ( ???? ) =
?? ?? (
?? ?? ?? ????
-
?? ?? ?? ????
) , ?? , ?? ? ?? , then
?? ( ?? + ?? ) is equal to
JEE Main 2025 (Online) 23rd January Morning Shift
Options:
A. 39
B. 22
C. 40
D. 26
Ans: A
Solution:
?? ( ?? ) = ?
????
( ?? - 11 )
11
13
( ?? + 15 )
15
13
Put
?? - 11
?? + 15
= ?? ?
26
( ?? + 5 )
2
???? = ????
?? ( ?? ) =
1
26
?
????
?? 11 / 13
=
1
26
·
?? 2 / 13
2 / 13
I ( x ) =
1
4
(
x - 11
x + 15
)
2 / 13
+ C
I ( 37 ) - I ( 24 ) =
1
4
(
26
52
)
2 /13
-
1
4
(
13
39
)
2 / 13
=
1
4
(
1
2
2 / 13
-
1
3
2 / 13
)
=
1
4
(
1
4
1 / 13
-
1
9
1 /13
)
? b = 4 , c = 9
3 ( b + c ) = 39
Q4: Let ? ?? ?? ?? ???? ? ?? ?? ?? = ?? ( ?? ) + ?? , where ?? is the constant of integration. If
?? ( ?? (
?? ?? ) + ?? '
(
?? ?? ) ) = ?? ?? ?? + ?? ?? ?? + ?? , ?? , ?? , ?? ? ?? , then ?? + ?? - ?? equals :
JEE Main 2025 (Online) 23rd January Evening Shift
Options:
A. 47
B. 55
C. 62
D. 48
Ans: B
Solution:
? ?? 3
sin ? ???? ?? = - ?? 3
c o s ? ?? + ? 3 ?? 2
c o s ? ???? ??
= - ?? 3
c os ? ?? + 3 ?? 2
s i n ? ?? - ? 6 ?? s i n ? ???? ??
= - ?? 3
c os ? ?? + 3 ?? 2
s i n ? ?? + 6 ?? c os ? ?? - 6s i n ? ?? + ??
So ?? ( ?? ) = - ?? 3
c os ? ?? + 3 ?? 2
s i n ? ?? + 6 ?? c os ? ?? - 6s i n ? ??
?? (
?? 2
) =
3 ?? 2
4
- 6
?? '
( ?? ) = ?? 3
s i n ? ??
?? '
(
?? 2
) =
?? 3
8
8 ( ?? (
?? 2
) + ?? '
(
?? 2
) ) = ?? 3
+ 6 ?? 2
- 48
Q5: If ?? ( ?? ) = ?
?? ?? ?? / ?? ( ?? + ?? ?? / ?? )
?? ?? , ?? ( ?? ) = - ?? , then ?? ( ?? ) is equal to :
JEE Main 2025 (Online) 28th January Evening Shift
Options:
A. 4 ( log
?? ? 2 - 2 )
B. log
?? 2 ? 2 + 2
C. 2 - log ? e
2
D. 4 ( log
?? ? 2 + 2 )
Ans: A
Solution:
Substitution: Start by substituting ?? = ?? 4
which implies ???? = 4 ?? 3
???? .
Integral Transformation:
?
1
?? 1 / 4
( 1 + ?? 1 / 4
)
???? ? ?
4 ?? 3
????
?? ( 1 + ?? )
= ?
4 ?? 1 + ?? ????
Rewriting the Integral: The expression
4 ?? 1 + ?? can be decomposed as:
4 ?? 1 + ?? = 4 (
?? 2
- 1 + 1
1 + ?? ) = 4 ( ( ?? - 1 ) +
1
?? + 1
)
Separate and Integrate:
4 ? ( ?? - 1 ) ???? + 4 ?
1
?? + 1
????
Solving the Integrals:
The integral of ?? - 1 is
( ?? - 1 )
2
2
.
The integral of
1
?? + 1
is ln ? ( ?? + 1 ) .
Page 5
JEE Main Previous Year Questions
(2025): Indefinite Integral
Q1: If ? ?? ?? (
?? ?? ?? ?? - ?? ? ?? v
?? - ?? ?? +
?? ?? ?? - ?? ? ?? ( ?? - ?? ?? )
?? / ?? +
?? ?? - ?? ?? ) ?? ?? = ?? ( ?? ) + ?? , where C is the constant of
integration, then ?? (
?? ?? ) equals :
JEE Main 2025 (Online) 8th April Evening Shift
Options:
A. 2 + 3 ??
B. 4 ? + ??
C. 1 + 3 ??
D. 3 + 2 ??
Ans: C
Solution:
Let, I = ?? 2
?
- 1
3 / 2
? | x s i n ? ?? x | dx
= ?? 2
{ ?
- 1
1
? x s i n ? ?? x dx - ?
1
3 / 2
? x s i n ? ?? x dx }
= ?? 2
{ 2 ?
0
1
? x s i n ? ?? x dx - ?
- 1
3 / 2
? x s i n ? ?? x dx }
Consider
? x s i n ? ?? x dx
- x ·
1
?? c os ? ?? x + ? 1 ·
1
?? c os ? ?? x dx
= - -
x
?? c os ? ?? x +
sin ? ?? x
?? 2
I = ?? 2
{ 2 ( -
x
?? c os ? ?? x +
sin ? ?? x
?? 2
)
0
1
- ( -
x
?? c os ? ?? x +
sin ? ?? x
?? 2
)
1
3 / 2
}
= ?? 2
{
2
?? - ( -
1
?? 2
-
1
?? ) }
= ?? 2
{
3
?? +
1
?? 2
}
= 3 ?? + 1
Q2: If ? ?? ?? (
?? ?? ???? - ?? ? ?? v
?? - ?? ?? +
?? ???? - ?? ? ?? ( ?? - ?? ?? )
?? / ?? +
?? ?? - ?? ?? ) ?? ?? = ?? ( ?? ) + ?? , where C is the constant of integration,
then ?? (
?? ?? ) equals :
Options:
A.
?? 6
v
e
3
B.
?? 6
v
e
2
C.
?? 4
v
e
3
D.
?? 4
v
e
2
Ans: A
Solution:
?
d
dx
(
x sin
- 1
? x
v 1 - x
2
) =
s i n
- 1
? x
( 1 - x
2
)
3 / 2
+
x
1 - x
2
? ? e
x
(
x sin
- 1
? x
v 1 - x
2
+
sin
- 1
? x
( 1 - x
2
)
3 / 2
+
x
1 - x
2
) dx
= e
x
·
x sin
- 1
? x
v 1 - x
2
+ c = g ( x ) + C
Note : assuming ?? ( ?? ) =
xe
x
s i n
- 1
? ?? v 1 - x
2
?? ( 1 / 2 ) =
e
1 / 2
2
·
?? 6
× 2
v 3
=
?? 6
v
e
3
Comment : In this Q we will not get a unique function g ( x ) , but in order to match the answer we
will have to assume ?? ( ?? ) =
xe s i n
- 1
? x
v 1 - x
2
.
Q3: Let ?? ( ?? ) = ?
????
( ?? - ???? )
????
???? ( ?? + ???? )
????
????
. If ?? ( ???? ) - ?? ( ???? ) =
?? ?? (
?? ?? ?? ????
-
?? ?? ?? ????
) , ?? , ?? ? ?? , then
?? ( ?? + ?? ) is equal to
JEE Main 2025 (Online) 23rd January Morning Shift
Options:
A. 39
B. 22
C. 40
D. 26
Ans: A
Solution:
?? ( ?? ) = ?
????
( ?? - 11 )
11
13
( ?? + 15 )
15
13
Put
?? - 11
?? + 15
= ?? ?
26
( ?? + 5 )
2
???? = ????
?? ( ?? ) =
1
26
?
????
?? 11 / 13
=
1
26
·
?? 2 / 13
2 / 13
I ( x ) =
1
4
(
x - 11
x + 15
)
2 / 13
+ C
I ( 37 ) - I ( 24 ) =
1
4
(
26
52
)
2 /13
-
1
4
(
13
39
)
2 / 13
=
1
4
(
1
2
2 / 13
-
1
3
2 / 13
)
=
1
4
(
1
4
1 / 13
-
1
9
1 /13
)
? b = 4 , c = 9
3 ( b + c ) = 39
Q4: Let ? ?? ?? ?? ???? ? ?? ?? ?? = ?? ( ?? ) + ?? , where ?? is the constant of integration. If
?? ( ?? (
?? ?? ) + ?? '
(
?? ?? ) ) = ?? ?? ?? + ?? ?? ?? + ?? , ?? , ?? , ?? ? ?? , then ?? + ?? - ?? equals :
JEE Main 2025 (Online) 23rd January Evening Shift
Options:
A. 47
B. 55
C. 62
D. 48
Ans: B
Solution:
? ?? 3
sin ? ???? ?? = - ?? 3
c o s ? ?? + ? 3 ?? 2
c o s ? ???? ??
= - ?? 3
c os ? ?? + 3 ?? 2
s i n ? ?? - ? 6 ?? s i n ? ???? ??
= - ?? 3
c os ? ?? + 3 ?? 2
s i n ? ?? + 6 ?? c os ? ?? - 6s i n ? ?? + ??
So ?? ( ?? ) = - ?? 3
c os ? ?? + 3 ?? 2
s i n ? ?? + 6 ?? c os ? ?? - 6s i n ? ??
?? (
?? 2
) =
3 ?? 2
4
- 6
?? '
( ?? ) = ?? 3
s i n ? ??
?? '
(
?? 2
) =
?? 3
8
8 ( ?? (
?? 2
) + ?? '
(
?? 2
) ) = ?? 3
+ 6 ?? 2
- 48
Q5: If ?? ( ?? ) = ?
?? ?? ?? / ?? ( ?? + ?? ?? / ?? )
?? ?? , ?? ( ?? ) = - ?? , then ?? ( ?? ) is equal to :
JEE Main 2025 (Online) 28th January Evening Shift
Options:
A. 4 ( log
?? ? 2 - 2 )
B. log
?? 2 ? 2 + 2
C. 2 - log ? e
2
D. 4 ( log
?? ? 2 + 2 )
Ans: A
Solution:
Substitution: Start by substituting ?? = ?? 4
which implies ???? = 4 ?? 3
???? .
Integral Transformation:
?
1
?? 1 / 4
( 1 + ?? 1 / 4
)
???? ? ?
4 ?? 3
????
?? ( 1 + ?? )
= ?
4 ?? 1 + ?? ????
Rewriting the Integral: The expression
4 ?? 1 + ?? can be decomposed as:
4 ?? 1 + ?? = 4 (
?? 2
- 1 + 1
1 + ?? ) = 4 ( ( ?? - 1 ) +
1
?? + 1
)
Separate and Integrate:
4 ? ( ?? - 1 ) ???? + 4 ?
1
?? + 1
????
Solving the Integrals:
The integral of ?? - 1 is
( ?? - 1 )
2
2
.
The integral of
1
?? + 1
is ln ? ( ?? + 1 ) .
Therefore:
?? ( ?? ) = 4 {
( ?? - 1 )
2
2
+ ln ? ( ?? + 1 ) } + ??
Substitute ?? = ?? 1 / 4
: Replace ?? back with ?? 1 / 4
:
?? ( ?? ) = 2 ( ?? 1 / 4
- 1 )
2
+ 4ln ? ( 1 + ?? 1 / 4
) + ??
Using the Condition ?? ( 0 ) = - 6 :
?? ( 0 ) = 2 ( 0 - 1 )
2
+ 4ln ? ( 1 + 0 ) + ?? = 2 × 1 - 0 + ?? = - 6
Solving gives ?? = - 8.
Find ?? ( 1 ) :
?? ( 1 ) = 2 ( 1
1 / 4
- 1 )
2
+ 4ln ? ( 1 + 1
1 / 4
) - 8
Simplifies to:
Q6: Let ?? ( ?? ) = ? ?? ?? v ?? - ?? ?? ???? . If ?? ?? ( v ?? ) = - ?? , then ?? ( ?? ) is equal to
JEE Main 2025 (Online) 3rd April Morning Shift
Options:
A. -
6 v 2
5
B. -
8 v 2
5
C. -
2 v 2
5
D. -
4 v 2
5
Ans: A
Solution:
? ? ? ?? 3
v
3 - ?? 2
????
3 - ?? 2
= ?? 2
? - 2 ???? ?? = 2 ??????
? = - ? ? ?? 2
( 3 - ?? 2
) ???? = ? ? ?? 4
- 3 ?? 2
????
? =
?? 5
5
- ?? 3
+ ??
?? ( ?? ) =
( 3 - ?? 2
)
5
2
5
- ( 3 - ?? 2
)
3
2
+ ??
? ?? ( v 2 ) = -
4
5
? ? -
4
5
=
1
5
- 1 + ?? ? ?? = 0
? ?? ( ?? ) =
( 3 - ?? 2
)
5
2
5
- ( 3 - ?? 2
)
3
2
Read MoreFAQs on JEE Main Previous Year Questions (2026): Indefinite Integral
| 1. What is the importance of indefinite integrals in calculus? |  |
Ans. Indefinite integrals are fundamental in calculus as they represent the family of antiderivatives of a function. They allow us to find functions whose derivatives yield the original function, which is essential for solving differential equations, calculating areas under curves, and understanding the accumulation of quantities.
| 2. How can one determine the indefinite integral of basic polynomial functions? | |
Ans. To find the indefinite integral of polynomial functions, apply the power rule. For a term in the form of ax^n, the integral is (a/n+1)x^(n+1) + C, where C is the constant of integration. For example, the integral of 3x² is (3/3)x³ + C, which simplifies to x³ + C.
| 3. What techniques can be used for integrating more complex functions? | |
Ans. For more complex functions, several techniques can be applied, including substitution, integration by parts, partial fraction decomposition, and trigonometric identities. Each method is suited for specific types of functions, helping to simplify the integration process.
| 4. How does the Fundamental Theorem of Calculus relate to indefinite integrals? | |
Ans. The Fundamental Theorem of Calculus connects differentiation and integration, stating that if F is an antiderivative of f on an interval [a, b], then the integral of f from a to b is F(b) - F(a). It emphasizes that indefinite integrals yield functions that can be evaluated over a range, reinforcing the concept of area under a curve.
| 5. What common mistakes should students avoid when solving indefinite integrals? | |
Ans. Common mistakes include forgetting to include the constant of integration, misapplying integration rules (such as power rule), and failing to simplify expressions before integrating. Additionally, students often overlook the need for substitution in composite functions, which can lead to incorrect results.
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