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JEE Advanced 2020 Full Test 6 Paper 2
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© examsnet.com
Question : 48
Total: 54
∫
x
2
−
1
x
3
√
2
x
4
−
2
x
2
+
1
d
x
is equal to
1
x
2
√
2
x
4
−
2
x
2
+
1
+
c
1
x
3
√
2
x
4
−
2
x
2
+
1
+
c
1
x
√
2
x
4
−
2
x
2
+
1
+
c
1
2
x
2
√
2
x
4
−
2
x
2
+
1
+
c
Validate
Solution:
∫
x
2
−
1
x
3
√
2
x
4
−
2
x
2
+
1
d
x
=
∫
x
2
−
1
x
5
√
2
−
2
x
2
+
1
x
4
d
x
=
∫
1
x
3
−
1
x
5
√
2
−
2
x
2
+
1
x
4
d
x
Let
2
−
2
x
2
+
1
x
4
=
t
2
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