integrate (x + 5)/(3x ^ 2 - 2x + 1) dx
Answers
Step-by-step explanation:
The answer is
=
x
2
2
−
3
4
ln
(
|
x
+
1
|
)
+
5
4
ln
(
|
x
−
1
|
)
−
1
4
ln
(
x
2
+
1
)
+
arctan
x
+
C
Explanation:
We need
∫
d
x
x
2
+
1
=
arctan
(
x
)
As the degree of the numerator is greater than the degree of the denominator, perform a long division first.
x
5
+
3
x
2
+
1
x
4
−
1
=
x
+
3
x
2
+
x
+
1
x
4
−
1
Perform a decomposition into partial fractions
3
x
2
+
x
+
1
x
4
−
1
=
3
x
2
+
x
+
1
(
x
2
+
1
)
(
x
+
1
)
(
x
−
1
)
=
A
x
+
B
x
2
+
1
+
C
x
+
1
+
D
x
−
1
=
(
A
x
+
B
)
(
x
+
1
)
(
x
−
1
)
+
C
(
x
2
+
1
)
(
x
−
1
)
+
D
(
x
2
+
1
)
(
x
+
1
)
(
x
2
+
1
)
(
x
+
1
)
(
x
−
1
)
The denominators are the same , compare the numerators
3
x
2
+
x
+
1
=
(
A
x
+
B
)
(
x
+
1
)
(
x
−
1
)
+
C
(
x
2
+
1
)
(
x
−
1
)
+
D
(
x
2
+
1
)
(
x
+
1
)
Let
x
=
1
,
⇒
,
5
=
4
D
,
⇒
,
D
=
5
4
Let
x
=
−
1
,
⇒
,
3
=
−
4
C
,
⇒
,
C
=
−
3
4
Let
x
=
0
,
⇒
, 1=-B-C+D
1
=
−
B
−
C
+
D
, =>
⇒
, B=5/4+3/4-1=1
B
=
5
4
+
3
4
−
1
=
1
Coefficients of x^3
x
3
0=A+C+D
0
=
A
+
C
+
D
, =>
⇒
, A=-C-D=3/4-5/4=-1/2
A
=
−
C
−
D
=
3
4
−
5
4
=
−
1
2
Therefore,
(x^5+3x^2+1)/(x^4-1)=x+(-1/2x+1)/(x^2+1)+(-3/4)/(x+1)+(5/4)/(x-1)
x
5
+
3
x
2
+
1
x
4
−
1
=
x
+
−
1
2
x
+
1
x
2
+
1
+
−
3
4
x
+
1
+
5
4
x
−
1
So,
int((x^5+3x^2+1)dx)/(x^4-1)=intxdx+int((-1/2x+1)dx)/(x^2+1)+int(-3/4dx)/(x+1)+int(5/4dx)/(x-1)
∫
(
x
5
+
3
x
2
+
1
)
d
x
x
4
−
1
=
∫
x
d
x
+
∫
(
−
1
2
x
+
1
)
d
x
x
2
+
1
+
∫
−
3
4
d
x
x
+
1
+
∫
5
4
d
x
x
−
1
intxdx=x^2/2
∫
x
d
x
=
x
2
2
int(-3/4dx)/(x+1)=-3/4ln(|x+1|)
∫
−
3
4
d
x
x
+
1
=
−
3
4
ln
(
|
x
+
1
|
)
int(5/4dx)/(x-1)=5/4ln(|x-1|)
∫
5
4
d
x
x
−
1
=
5
4
ln
(
|
x
−
1
|
)
int((-1/2x+1)dx)/(x^2+1)=-1/4int(2xdx)/(x^2+1)+intdx/(x^2+1)=-1/4ln(x^2+1)+arctanx
∫
(
−
1
2
x
+
1
)
d
x
x
2
+
1
=
−
1
4
∫
2
x
d
x
x
2
+
1
+
∫
d
x
x
2
+
1
=
−
1
4
ln
(
x
2
+
1
)
+
arctan
x
Finally,
int((x^5+3x^2+1)dx)/(x^4-1)=x^2/2-3/4ln(|x+1|)+5/4ln(|x-1|)-1/4ln(x^2+1)+arctanx+C