evaluate 1+x+√x+x^2/√x+√1+x.dx
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Answer:
d
x
x
+
√
x
2
+
x
+
1
=
(
√
x
2
+
x
+
1
−
x
)
+
ln
|
x
+
1
|
−
1
2
ln
∣
∣
2
x
+
1
+
2
√
x
2
+
x
+
1
∣
∣
+
ln
∣
∣
∣
∣
∣
(
2
−
√
3
)
(
√
3
+
2
√
x
2
+
x
+
1
)
+
(
2
x
+
1
)
(
2
−
√
3
)
(
√
3
+
2
√
x
2
+
x
+
1
)
−
(
2
x
+
1
)
∣
∣
∣
∣
∣
Explanation:
Let
I
=
∫
d
x
x
+
√
x
2
+
x
+
1
Rationalize:
I
=
∫
d
x
x
+
√
x
2
+
x
+
1
⋅
x
−
√
x
2
+
x
+
1
x
−
√
x
2
+
x
+
1
Simplify:
I
=
∫
√
x
2
+
x
+
1
−
x
x
+
1
d
x
Rearrange:
I
=
∫
√
x
2
+
x
+
1
x
+
1
d
x
−
∫
(
1
−
1
x
+
1
)
d
x
Complete the square in the square root:
I
=
1
2
∫
√
(
2
x
+
1
)
2
+
3
x
+
1
d
x
−
(
x
−
ln
|
x
+
1
|
)
Apply the substitution
2
x
+
1
=
√
3
tan
θ
:
I
=
1
2
∫
√
3
sec
θ
√
3
2
tan
θ
+
1
2
(
√
3
2
sec
2
θ
d
θ
)
−
x
+
ln
|
x
+
1
|
Simplify:
I
=
3
2
∫
sec
2
θ
√
3
tan
θ
+
1
sec
θ
d
θ
−
x
+
ln
|
x
+
1
|
Since
sec
2
θ
=
tan
2
θ
+
1
:
I
=
1
2
∫
(
√
3
tan
θ
−
1
+
4
√
3
tan
θ
+
1
)
sec
θ
d
θ
−
x
+
ln
|
x
+
1
|
Rearrange:
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