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Exams
Joint Entrance Examination
Mathematics
Indefinite Integrals
Question 224
Question 224 of 416
Medium
MCQ
The integral
∫
1
2
e
x
⋅
x
x
(
2
+
log
e
x
)
d
x
\displaystyle\int_1^2e^x\cdot x^x(2+\log_ex)\,dx
∫
1
2
e
x
⋅
x
x
(
2
+
lo
g
e
x
)
d
x
equals:
A
e
(
4
e
+
1
)
e(4e+1)
e
(
4
e
+
1
)
B
4
e
2
−
1
4e^2-1
4
e
2
−
1
C
e
(
4
e
−
1
)
e(4e-1)
e
(
4
e
−
1
)
D
e
(
2
e
−
1
)
e(2e-1)
e
(
2
e
−
1
)
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Solutions · 1
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Official solution
\[I=\int_1^2e^xx^x[1+(1+\log_ex)]\,dx=\int_1^2e^x\left[x^x+x^x(1+\log_ex)\right]dx\] Since
d
d
x
(
x
x
)
=
x
x
(
1
+
ln
x
)
\dfrac{d}{dx}(x^x)=x^x(1+\ln x)
d
x
d
(
x
x
)
=
x
x
(
1
+
ln
x
)
, using
∫
e
x
(
f
(
x
)
+
f
′
(
x
)
)
d
x
=
e
x
f
(
x
)
+
c
\displaystyle\int e^x(f(x)+f'(x))dx=e^xf(x)+c
∫
e
x
(
f
(
x
)
+
f
′
(
x
))
d
x
=
e
x
f
(
x
)
+
c
with
f
(
x
)
=
x
x
f(x)=x^x
f
(
x
)
=
x
x
: \[I=\left[e^xx^x\right]_1^2=e^2\cdot4-e\cdot1=4e^2-e=e(4e-1)\]
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