integration of x³/((x-1)*(x²+1))
Answers
Let =∫3(−1)(2+1)
I
=
∫
x
3
(
x
−
1
)
(
x
2
+
1
)
d
x
=∫3−1+1(−1)(2+1)
=
∫
x
3
−
1
+
1
(
x
−
1
)
(
x
2
+
1
)
d
x
=∫(−1)(2++1)(−1)(2+1)+∫1(−1)(2+1)1
=
∫
(
x
−
1
)
(
x
2
+
x
+
1
)
(
x
−
1
)
(
x
2
+
1
)
d
x
+
∫
1
(
x
−
1
)
(
x
2
+
1
)
d
x
⏟
I
1
=∫2++12+1+1
=
∫
x
2
+
x
+
1
x
2
+
1
d
x
+
I
1
=∫+∫2+12+1
=
∫
d
x
+
∫
x
x
2
+
1
d
x
⏟
I
2
+
I
1
=+2+1
=
x
+
I
2
+
I
1
Lets solve the easiest first
So, 2=∫22(2+1)
I
2
=
∫
2
x
2
(
x
2
+
1
)
d
x
=∫(2+1)2(2+1)
=
∫
d
(
x
2
+
1
)
2
(
x
2
+
1
)
=12ln(2+1)
=
1
2
ln
(
x
2
+
1
)
Now lets solve 1
I
1
,
1=∫1(−1)(2+1)
I
1
=
∫
1
(
x
−
1
)
(
x
2
+
1
)
d
x
Lets apply the partial fraction decomposition technique
Assuming 1(−1)(2+1)=−1++2+1——(1)
1
(
x
−
1
)
(
x
2
+
1
)
=
A
x
−
1
+
B
x
+
C
x
2
+
1
—
—
(
1
)
⟹(2+1)+(+)(−1)=1
⟹
A
(
x
2
+
1
)
+
(
B
x
+
C
)
(
x
−
1
)
=
1
At =1
x
=
1
2=1
2
A
=
1
⟹=12
⟹
A
=
1
2
At ==−1‾‾‾√
x
=
i
=
−
1
(+)(−1)=1
(
B
i
+
C
)
(
i
−
1
)
=
1
−+−−=1
−
B
+
C
i
−
B
i
−
C
=
1
⟹−=0
⟹
C
−
B
=
0
and +=−1
B
+
C
=
−
1
After solving above two linear equation, we get
=−12
C
=
−
1
2
=−12
B
=
−
1
2
After putting value of
A
,
B
, and
C
into equation (1)
(
1
)
and then replacing this partial fraction into 1
I
1
, we get,
1=∫12(−1)−∫+12(2+1)
I
1
=
∫
1
2
(
x
−
1
)
d
x
−
∫
x
+
1
2
(
x
2
+
1
)
d
x
=12ln(−1)−∫2(2+1)−∫12(2+1)
=
1
2
ln
(
x
−
1
)
−
∫
x
2
(
x
2
+
1
)
d
x
−
∫
1
2
(
x
2
+
1
)
d
x
=12ln(−1)−∫24(2+1)−12arctan()
=
1
2
ln
(
x
−
1
)
−
∫
2
x
4
(
x
2
+
1
)
d
x
−
1
2
arctan
(
x
)
=12ln(−1)−∫(2+1)4(2+1)−12arctan()
=
1
2
ln
(
x
−
1
)
−
∫
d
(
x
2
+
1
)
4
(
x
2
+
1
)
−
1
2
arctan
(
x
)
=12ln(−1)−14ln(2+1)−12arctan()
=
1
2
ln
(
x
−
1
)
−
1
4
ln
(
x
2
+
1
)
−
1
2
arctan
(
x
)
Now putting the values of 1
I
1
and 2
I
2
into
I
, we get,
=+12ln(2+1)+12ln(−1)−14ln(2+1)−12arctan()+
I
=
x
+
1
2
ln
(
x
2
+
1
)
+
1
2
ln
(
x
−
1
)
−
1
4
ln
(
x
2
+
1
)
−
1
2
arctan
(
x
)
+
C
(
C
is the constant of integration)