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Byju's Answer
Standard XII
Mathematics
Property 7
integration o...
Question
integration of x
2
+1/x(x
2
-1) by partial fraction
Open in App
Solution
∫
x
2
+
1
x
x
2
-
1
dx
=
∫
x
2
+
1
x
x
+
1
x
-
1
dx
Let
x
2
+
1
x
x
+
1
x
-
1
=
A
x
+
B
x
+
1
+
C
x
-
1
x
2
+
1
x
x
+
1
x
-
1
=
A
x
+
1
x
-
1
+
Bx
x
-
1
+
Cx
x
+
1
x
x
+
1
x
-
1
x
2
+
1
x
x
+
1
x
-
1
=
A
x
2
-
1
+
B
x
2
-
x
+
C
x
2
+
x
x
x
+
1
x
-
1
x
2
+
1
x
x
+
1
x
-
1
=
A
+
B
+
C
x
2
+
-
B
+
C
x
-
A
x
x
+
1
x
-
1
On
comparing
the
coefficients
,
we
get
A
+
B
+
C
=
1
.
.
.
1
-
B
+
C
=
0
.
.
.
2
A
=
-
1
.
.
.
3
On
solving
1
,
2
and
3
,
we
get
A
=
-
1
,
B
=
1
and
C
=
1
So
,
∫
x
2
+
1
x
x
2
-
1
dx
=
-
∫
1
x
dx
+
∫
1
x
+
1
dx
+
∫
1
x
-
1
dx
=
-
log
x
+
log
x
+
1
+
log
x
-
1
+
c
=
log
x
2
-
1
-
log
x
+
c
=
log
x
2
-
1
x
+
c
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0
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Q.
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. Find the sum of coefficients of the individual terms.
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