6whole 1/2 + 1 whole 1 upon 2 (1 - 2 X) =_4
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x
−
arctan
x
+
C
Explanation:
x
2
x
2
+
1
=
x
2
+
1
−
1
x
2
+
1
=
1
−
1
x
2
+
1
∫
1
−
1
x
2
+
1
d
x
=
x
−
∫
1
x
2
+
1
d
x
in terms of the red bit, use sub
x
=
tan
t
,
d
x
=
sec
2
t
d
t
this makes it
∫
1
tan
2
t
+
1
sec
2
t
d
t
=
∫
1
sec
2
t
sec
2
t
d
t
=
∫
d
t
=
arctan
x
−
C
So the full integral is
x
−
arctan
x
+
C
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