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Byju's Answer
Standard XII
Mathematics
Factorization
Evaluate the ...
Question
Evaluate the following integrals:
∫
1
4
x
-
1
+
x
-
2
+
x
-
4
d
x
Open in App
Solution
I
=
∫
1
4
x
-
1
+
x
-
2
+
x
-
4
d
x
⇒
I
=
∫
1
4
x
-
1
d
x
+
∫
1
4
x
-
2
d
x
+
∫
1
4
x
-
4
d
x
We
know
that
,
x
-
1
=
-
x
-
1
,
x
≤
1
x
-
1
,
1
<
x
≤
4
x
-
2
=
-
x
-
2
,
1
≤
x
≤
2
x
-
2
,
2
<
x
≤
4
x
-
4
=
-
x
-
4
,
1
≤
x
≤
4
x
-
4
,
x
>
4
∴
I
=
∫
1
4
x
-
1
d
x
-
∫
1
2
x
-
2
d
x
+
∫
2
4
x
-
2
d
x
-
∫
1
4
x
-
4
d
x
⇒
I
=
x
2
2
-
x
1
4
-
x
2
2
-
2
x
1
2
+
x
2
2
-
2
x
2
4
-
x
2
2
-
4
x
1
4
⇒
I
=
8
-
4
-
1
2
+
1
-
2
-
4
-
1
2
+
2
+
8
-
8
-
2
+
4
-
8
-
16
-
1
2
+
4
⇒
I
=
23
2
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