We will use integration by substitution, as well as the integrals
∫
1
x
d
x
=
ln
|
x
|
+
C
and
∫
1
d
x
=
x
+
C
∫
x
3
x
2
+
1
d
x
=
∫
x
2
x
2
+
1
x
d
x
=
1
2
∫
(
x
2
+
1
)
−
1
x
2
+
1
2
x
d
x
Let
u
=
x
2
+
1
⇒
d
u
=
2
x
d
x
. Then
1
2
∫
(
x
2
+
1
)
−
1
x
2
+
1
2
x
d
x
=
1
2
∫
u
−
1
u
d
u
=
1
2
∫
(
1
−
1
u
)
d
u
=
1
2
(
u
−
ln
|
u
|
)
+
C
=
x
2
+
1
2
−
ln
(
x
2
+
1
)
2
+
C
=
x
2
2
−
ln
(
x
2
+
1
)
2
+
1
2
+
C
=
x
2
−
ln
(
x
2
+
1
)
2
+
C
Final answer
The following figures are similar. They are both the same shape, One is just bigger.
Answer:
do we have to put an whole number there
Answer:
(p,r) = (1/3, 2/9)
Step-by-step explanation:
Here, we want to solve a system of equations
We can rewrite the second equation by dividing through by 2
So we have;
4p + 3r = 2
and
5p - 3r = 1
Add both equations:
9p = 3
p = 3/9
p = 1/3
Recall ;
5p - 3r = 1
3r = 5p - 1
Substitute the value or p here
3r = 5(1/3)-1
3r = 5/3 - 1
3r = 2/3
r = 2/9
So we have the solution set as;
(p,r) = (1/3 , 2/9)
Answer:
-6M
Step-by-step explanation:
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