integration of x^3+1/x^2+1
Answers
Answer:
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Calculus Examples
Popular Problems Calculus Evaluate integral of (x^3+1)/(x^2-1) with respect to x
∫
x
3
+
1
x
2
−
1
d
x
Simplify.
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∫
x
2
−
x
+
1
x
−
1
d
x
Divide
x
2
−
x
+
1
by
x
−
1
.
∫
x
+
1
x
−
1
d
x
Split the single integral into multiple integrals.
∫
x
d
x
+
∫
1
x
−
1
d
x
By the Power Rule, the integral of
x
with respect to
x
is
1
2
x
2
.
1
2
x
2
+
C
+
∫
1
x
−
1
d
x
Let
u
=
x
−
1
. Then
d
u
=
d
x
. Rewrite using
u
and
d
u
.
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1
2
x
2
+
C
+
∫
1
u
d
u
The integral of
1
u
with respect to
u
is
ln
(
|
u
|
)
.
1
2
x
2
+
C
+
ln
(
|
u
|
)
+
C
Simplify.
1
2
x
2
+
ln
(
|
u
|
)
+
C
Replace all occurrences of
u
with
x
−
1
.
1
2
x
2
+
ln
(
|
x
−
1
|
)
+
C
Answer:
Simplify.
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∫
x
2
−
x
+
1
x
−
1
d
x
∫x2-x+1x-1dx
Divide
x
2
−
x
+
1
x2-x+1
by
x
−
1
x-1
.
∫
x
+
1
x
−
1
d
x
∫x+1x-1dx
Split the single integral into multiple integrals.
∫
x
d
x
+
∫
1
x
−
1
d
x
∫xdx+∫1x-1dx
By the Power Rule, the integral of
x
x
with respect to
x
x
is
1
2
x
2
12x2
.
1
2
x
2
+
C
+
∫
1
x
−
1
d
x
12x2+C+∫1x-1dx
Let
u
=
x
−
1
u=x-1
. Then
d
u
=
d
x
du=dx
. Rewrite using
u
u
and
d
d
u
u
.
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1
2
x
2
+
C
+
∫
1
u
d
u
12x2+C+∫1udu
The integral of
1
u
1u
with respect to
u
u
is
ln
(
|
u
|
)
ln(|u|)
.
1
2
x
2
+
C
+
ln
(
|
u
|
)
+
C
12x2+C+ln(|u|)+C
Simplify.
1
2
x
2
+
ln
(
|
u
|
)
+
C
12x2+ln(|u|)+C
Replace all occurrences of
u
u
with
x
−
1
x-1
.
1
2
x
2
+
ln
(
|
x
−
1
|
)
+
C
12x2+ln(|x-1|)+C
OR
Explanation:
First note that the given integrand (function to be integrated) is an Improper Rational Function [degree of poly. in Numerator (Nr.)
=
3
>
2
=
degree of poly. in Deno.(Dr.)]. So, before integrating, we have to make it Proper. Usually, Long Division is performed for this, but we proceed as under:
N
r
.
=
x
3
=
x
3
+
x
−
x
=
x
(
x
2
+
1
)
−
x
.
∴
Integrand = Nr./Dr. =
x
(
x
2
+
1
)
−
x
x
2
+
1
=
x
(
x
2
+
1
)
x
2
+
1
−
x
x
2
+
1
=
x
−
x
x
2
+
1
∴
∫
x
3
x
2
+
1
d
x
=
∫
[
x
−
x
x
2
+
1
]
d
x
=
∫
x
d
x
−
(
1
2
)
∫
2
x
x
2
+
1
d
x
=
x
2
2
−
1
2
ln
(
x
2
+
1
)
+
C
.
Notice that the second integral follows by using the formula ;
∫
f
'
(
x
)
f
(
x
)
d
x
=
ln
f
(
x
)
Step-by-step explanation:
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