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
Answer:
128
Step-by-step explanation:
sorry if this is wrong im just too bored
Answer:
12
Step-by-step explanation:
Answer:
see below
Step-by-step explanation:
The exponent rules that apply are ...
(a^b)(a^c) = a^(b+c)
a^-b = (1/a)^b
(a^b)^c = a^(b·c)
_____
These let you rewrite the given function as ...
f(x) = (3^(2x))(3^1) = 3(3^(2x)) = 3(3^2)^x = 3·9^x
and
f(x) = 3^(2x+1) = (3^-1)^(-(2x+1)) = (1/3)^-(2x+1)
Explain yourself more and I’ll help