integration (1+X)^2/√x DX
Answers
Answered by
1
Step-by-step explanation:
Let
I
=
∫
1
x
2
+
x
d
x
∴
I
=
∫
1
x
(
x
+
1
)
d
x
=
∫
(
x
+
1
)
−
x
x
(
x
+
1
)
d
x
=
∫
{
x
+
1
x
(Let
I
=
∫
1
x
2
+
x
d
x
∴
I
=
∫
1
x
(
x
+
1
)
d
x
=
∫
(
x
+
1
)
−
x
x
(
x
+
1
)
d
x
=
∫
{
x
+
1
x
(
x
+
1
)
−
x
x
(
x
+
1
)
}
d
x
=
∫
{
1
x
−
1
x
+
1
}
d
x
=
ln
|
x
|
−
ln
|
x
+
1
|
=
ln
∣
∣
∣
x
x
+
1
∣
∣
∣
+
C
x
+
1
)
−
x
x
(
x
+
1
)
}
d
x
=
∫
{
1
x
−
1
x
+
1
}
d
x
=
ln
|
x
|
−
ln
|
x
+
1
|
=
ln
∣
∣
∣
x
x
+
1
∣
∣
∣
+
C
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