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Step-by-step explanation:
∫
x
3
2x
4
−2x
2
+1
x
2
−1
dx
⇒ ∫
x
5
x
3
2x
4
−2x
2
+1
x
5
x
2
−1
dx
⇒ ∫
x
2
2x
4
−2x
2
+1
x
3
1
−
x
5
1
dx
⇒ ∫
2−
x
2
2
+
x
4
1
x
3
1
−
x
5
1
dx
Let 2−
x
2
2
+
x
4
1
=t
(0+
x
3
4
−
x
5
4
)dx=dt
(
x
3
1
−
x
5
1
)=
4
dt
Now, substituting above value,
⇒ ∫
4
t
dt
⇒
4
1
∫t
2
−1
dt
⇒
4
1
2
−1
+1
t
2
−1
+1
+c
⇒
4
1
2
1
t
2
1
+c
⇒
2
t
+c
⇒
2
2−
x
2
2
+
x
4
1
+c
⇒
2x
2
2x
4
−2x
2
+1
+c
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