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Byju's Answer
Standard XII
Mathematics
Integration Using Substitution
If ∫ln 2k1/√e...
Question
If
k
∫
ln
2
1
√
e
x
−
1
d
x
=
π
6
,
then
k
=
ln
p
.
The value of
p
+
1
is
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Solution
k
∫
ln
2
1
√
e
x
−
1
d
x
Let
e
x
−
1
=
t
2
⇒
e
x
d
x
=
2
t
d
t
⇒
d
x
=
2
t
t
2
+
1
d
t
∴
∫
1
√
e
x
−
1
d
x
=
∫
2
t
2
+
1
d
t
=
2
tan
−
1
t
Now,
k
∫
ln
2
1
√
e
x
−
1
d
x
=
π
6
⇒
2
tan
−
1
√
e
k
−
1
−
2
tan
−
1
√
e
ln
2
−
1
=
π
6
⇒
2
tan
−
1
√
e
k
−
1
−
2
tan
−
1
1
=
π
6
⇒
2
tan
−
1
√
e
k
−
1
−
π
2
=
π
6
⇒
tan
−
1
√
e
k
−
1
=
π
3
⇒
√
e
k
−
1
=
√
3
⇒
e
k
=
4
⇒
k
=
ln
4
Thus
p
=
4
⇒
p
+
1
=
5
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