Question

1)S¹(x+3)dx=
2)S ³(x²+x+1)dx=
3)S⁴(2x+5)dx=
4)S¹(x-3)dx
5)S¹(x³+x+2)dx
6S¹(4×+3)dx
​

Answer 1

We know that
(
1
)
∫
f
′
(
x
)
f
(
x
)
d
x
=
ln
|
f
(
x
)
|
+
c

(
2
)
∫
1
U
2
+
1
d
x
=
tan
−
1
U
+
c

Now
I
=
∫
2
x
−
5
x
2
+
2
x
+
2
d
x

=
∫
2
x
+
2
−
7
x
2
+
2
x
+
2
d
x

=
∫
2
x
+
2
x
2
+
2
x
+
2
d
x
−
∫
7
x
2
+
2
x
+
2
d
x

=
∫
d
d
x
(
x
2
+
2
x
+
2
)
x
2
+
2
x
+
2
d
x
−
7
∫
1
x
2
+
2
x
+
1
+
1
d
x
I
=
∫
d
d
x
(
x
2
+
2
x
+
2
)
x
2
+
2
x
+
2
d
x
−
7
∫
1
(
x
+
1
)
2
+
1
d
x
Using
(
1
)
and
(
2
)
,
we get
I
=
ln
∣
∣
x
2
+
2
x
+
2
∣
∣
−
7
tan
−
1
(
x
+
1
)
+
c