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Byju's Answer
Standard XII
Mathematics
Particular Solution of a Differential Equation
Solve : ∫x ...
Question
Solve :
∫
x
d
x
(
x
2
+
a
2
)
(
x
2
+
b
2
)
Open in App
Solution
∫
x
(
x
2
+
a
2
)
(
x
2
+
b
2
)
d
x
substitute
u
=
x
2
2
→
d
x
=
1
x
d
u
=
∫
1
(
2
u
+
a
2
)
(
2
u
+
b
2
)
d
u
Upon partial fraction decomposition
=
∫
(
1
(
b
2
−
a
2
)
(
2
u
+
a
2
)
−
1
(
b
2
−
a
2
)
(
2
u
+
b
2
)
)
d
u
upon linearity
=
1
(
b
2
−
a
2
)
∫
1
2
u
+
a
2
d
u
−
1
(
b
2
−
a
2
)
∫
1
2
u
+
b
2
d
u
=
1
(
b
2
−
a
2
)
ln
(
2
u
+
a
2
)
2
−
1
(
b
2
−
a
2
)
ln
(
2
u
+
b
2
)
2
=
−
ln
(
x
2
+
b
2
)
−
ln
(
x
2
+
a
2
)
2
(
b
2
−
a
2
)
+
C
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Similar questions
Q.
∫
x
d
x
(
x
2
+
a
2
)
(
x
2
+
b
2
)
Q.
Solve
∫
1
(
x
2
+
a
2
)
(
x
2
+
b
2
)
d
x
.
Q.
Solve:
∫
∞
0
1
(
x
2
+
a
2
)
(
x
2
+
b
2
)
d
x
Q.
Solve the following quadratic equation using quadratic formula:
a
2
x
2
+
(
a
2
−
b
2
)
x
−
b
2
=
0
Q.
x
2
(
x
2
+
a
2
)
(
x
2
+
b
2
)
=
k
[
a
2
x
2
+
a
2
−
b
2
x
2
+
b
2
]
⇒
k
=
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Standard XII Mathematics
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