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AP EAMCET Engineering 21 Apr 2019 Shift 2 Paper
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© examsnet.com
Question : 74
Total: 160
∫
d
x
(
x
+
1
)
2
(
x
2
+
1
)
=
log
e
√
x
+
1
+
1
2
log
e
√
x
2
+
1
-
1
x
+
1
+ c
log
e
√
x
+
1
-
1
2
log
e
√
x
2
+
1
-
1
2
(
x
+
1
)
+ c
1
2
log
e
√
x
+
1
-
1
4
log
e
√
x
2
+
1
+
1
2
(
x
+
1
)
+ c
1
4
log
e
√
x
+
1
+
1
2
log
e
√
x
2
+
1
-
1
x
+
1
+ c
Validate
Solution:
The integral expression is given as,
I
=
∫
d
x
(
x
+
1
)
2
(
x
2
+
1
)
Consider
1
(
x
+
1
)
2
(
x
2
+
1
)
=
A
(
x
+
1
)
2
+
B
(
x
2
+
1
)
+
C
x
+
D
x
2
+
1
1
=
A
(
x
2
+
1
)
+
B
(
x
+
1
)
(
x
2
+
1
)
+
(
C
x
+
D
)
(
x
+
1
)
2
Substitute
x
=
−
1
A
=
1
2
Substitute
x
2
=
−
1
1
=
(
C
x
+
D
)
(
x
2
+
2
x
+
1
)
1
=
(
C
x
+
D
)
2
x
1
=
−
2
C
+
2
D
x
Substitute
x
=
0
−
1
2
=
C
,
D
=
0
Comparing coefficient of
x
3
0
=
B
+
C
B
=
1
2
Now
I
=
∫
[
1
2
1
(
x
+
1
)
2
+
1
2
1
x
+
1
−
1
2
x
x
2
+
1
]
d
x
=
1
2
[
−
1
x
+
1
+
log
(
x
+
1
)
−
1
2
log
(
x
2
+
1
)
]
+
c
=
log
√
x
+
1
−
1
2
log
√
x
2
+
1
−
1
2
(
x
+
1
)
+
c
© examsnet.com
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