evaluate integration x^2+2x+3/x^2 +x+1
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Answer:
Let's factorise the denominator
x
2
−
2
x
−
3
=
(
x
+
1
)
(
x
−
3
)
Let's perform the decomposition into partial fractions
x
+
2
(
x
+
1
)
(
x
−
3
)
=
A
x
+
1
+
B
x
−
3
=
A
(
x
−
3
)
+
B
(
x
+
1
)
(
x
+
1
)
(
x
−
3
)
The denominators are the same, we can equalize the numerators
(
x
+
2
)
=
A
(
x
−
3
)
+
B
(
x
+
1
)
Let
x
=
−
1
,
⇒
,
1
=
−
4
A
,
⇒
,
A
=
−
1
4
Let
x
=
3
,
⇒
,
5
=
4
B
,
B
=
5
4
Therefore,
x
+
2
(
x
+
1
)
(
x
−
3
)
=
−
1
4
x
+
1
+
5
4
x
−
3
∫
(
x
+
2
)
d
x
(
x
+
1
)
(
x
−
3
)
=
−
1
4
∫
d
x
x
+
1
+
5
4
∫
d
x
x
−
3
=
−
1
4
ln
(
|
x
+
1
|
)
+
5
4
ln
(
|
x
−
3
|
)
+
C
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