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Standard XII
Mathematics
RS Agarwal
Question
Evaluate the integral
∫
1
0
x
x
2
+
1
d
x
using substitution.
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Solution
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∫
1
0
x
x
2
+
1
d
x
Let
x
2
+
1
=
t
⇒
2
x
d
x
=
d
t
When
x
=
0
,
t
=
1
and when
x
=
1
,
t
=
2
∴
∫
1
0
x
x
2
+
1
d
x
=
1
2
∫
2
1
d
t
t
=
1
2
[
log
|
t
|
]
2
1
=
1
2
[
log
2
−
log
1
]
=
1
2
log
2
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