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:
If I've done it right the answer should be A, the figures are congruent because a 270 rotation about the origin a d a reflection of the x-axis
Answer:
(ab - 6)(2ab + 5)
Step-by-step explanation:
Assuming you require the expression factorised.
2a²b² - 7ab - 30
Consider the factors of the product of the coefficient of the a²b² term and the constant term which sum to give the coefficient of the ab- term
product = 2 × - 30 = - 60 and sum = - 7
The factors are - 12 and + 5
Use these factors to split the ab- term
= 2a²b² - 12ab + 5ab - 30 ( factor the first/second and third/fourth terms )
= 2ab(ab - 6) + 5(ab - 6) ← factor out (ab - 6) from each term
= (ab - 6)(2ab + 5) ← in factored form
2/1
Step-by-step explanation:
_1/3y=_1
y=_1+3/1
y=2/1
