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Byju's Answer
Standard XI
Mathematics
Integration by Substitution
The evaluatio...
Question
The evaluation of
∫
p
x
p
+
2
q
−
1
−
q
x
q
−
1
x
2
p
+
2
q
+
2
x
p
+
q
+
1
d
x
is
A
−
x
p
x
p
+
q
+
1
+
C
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B
x
q
x
p
+
q
+
1
+
C
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C
−
x
q
x
p
+
q
+
1
+
C
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D
x
p
x
p
+
q
+
1
+
C
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Solution
The correct option is
C
−
x
q
x
p
+
q
+
1
+
C
I
=
∫
p
x
p
+
2
q
−
1
−
q
x
q
−
1
(
x
p
+
q
+
1
)
2
d
x
=
∫
p
x
p
−
1
−
q
x
−
q
−
1
(
x
p
+
x
−
q
)
2
d
x
Put
x
p
+
x
−
q
=
t
p
x
p
−
1
−
q
x
−
q
−
1
d
x
=
d
t
Solving we get I =
−
x
q
x
p
+
q
+
1
+
C
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3
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Q.
∫
p
x
p
+
2
q
−
1
−
q
x
q
−
1
x
2
p
+
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q
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+
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2
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q
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+
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×
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(
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q
−
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÷
(
4
q
)
1
/
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×
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and find the negative of power of
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Q.
Every even integer and odd integer are of the form of
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and
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respectively, where
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