dx/ x⁴+x⁶ integrate the following
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Answer:
I=∫
x
6
+1
x
4
+1
dx
=∫
(x
6
+1)
(x
4
+1)
×
(x
2
+1)
(x
2
+1)
dx
=∫
(x
6
+1)(x
2
+1)
x
6
+x
2
+x
4
+⊥
dx
=∫
(x
6
+1)(x
2
+1)
(x
6
+1)+x
2
(x
2
+1)
dx
=∫
(x
6
+1)(x
2
+1)
(x
6
+1)
dx+∫
(x
6
+1)(x
2
+1)
x
2
(x
2
+1)
dx
=∫
(x
2
+1)
dx
+
3
1
∫
(x
6
+1)
3x
2
dx
Put, x
3
=t
⇒3x
2
dx=dt
=∫
(x
2
+1)
dx
+
3
1
∫
(x
3
)
2
+1
3x
2
dx
=∫
(x
2
+1)
dx
+
3
1
∫
t
2
+1
dt
=tan
−1
x+
3
1
tan
−1
(t)+c
=tan
−1
(x)+
3
1
tan
−1
(x
3
)+c.
Step-by-step explanation:
Attachments:
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