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Byju's Answer
Standard XII
Mathematics
Special Integrals - 2
Evaluate the ...
Question
Evaluate the following integrals:
∫
x
2
+
x
+
1
x
2
+
1
x
+
2
d
x
Open in App
Solution
Let
I
=
∫
x
2
+
x
+
1
x
2
+
1
x
+
2
d
x
We
express
x
2
+
x
+
1
x
2
+
1
x
+
2
=
A
x
+
2
+
B
x
+
C
x
2
+
1
⇒
x
2
+
x
+
1
=
A
x
2
+
1
+
B
x
+
C
x
+
2
Equating
the
coefficients
of
x
2
,
x
and
constants
,
we
get
1
=
A
+
B
and
1
=
2
B
+
C
and
1
=
A
+
2
C
or
A
=
3
5
and
B
=
2
5
and
C
=
1
5
∴
I
=
∫
3
5
x
+
2
+
2
5
x
+
1
5
x
2
+
1
d
x
=
3
5
∫
1
x
+
2
d
x
+
2
5
∫
x
x
2
+
1
d
x
+
1
5
∫
1
x
2
+
1
d
x
=
3
5
I
1
+
2
5
I
2
+
1
5
I
3
.
.
.
(
1
)
Now
,
I
1
=
∫
1
x
+
2
d
x
Let
x
+
2
=
u
On
differentiating
both
sides
,
we
get
d
x
=
d
u
∴
I
1
=
∫
1
u
d
u
=
log
u
+
c
1
=
log
x
+
2
+
c
1
.
.
.
(
2
)
And
,
I
2
=
∫
x
x
2
+
1
d
x
Let
x
2
+
1
=
u
On
differentiating
both
sides
,
we
get
2
x
d
x
=
d
u
∴
I
2
=
1
2
∫
1
u
d
u
=
1
2
log
u
+
c
2
=
1
2
log
x
2
+
1
+
c
2
.
.
.
(
3
)
And
,
I
3
=
∫
1
x
2
+
1
d
x
=
tan
-
1
x
+
c
3
.
.
.
(
4
)
From
(
1
)
,
(
2
)
,
(
3
)
and
(
4
)
,
we
get
∴
I
=
3
5
log
x
+
2
+
c
1
+
2
5
1
2
log
x
2
+
1
+
c
2
+
1
5
tan
-
1
x
+
c
3
=
3
5
log
x
+
2
+
1
5
log
x
2
+
1
+
1
5
tan
-
1
x
+
c
Hence
,
∫
x
2
+
x
+
1
x
2
+
1
x
+
2
d
x
=
3
5
log
x
+
2
+
1
5
log
x
2
+
1
+
1
5
tan
-
1
x
+
c
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