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Joint Entrance Examination
Mathematics
Indefinite Integrals
Question 406
Question 406 of 416
Hard
MCQ
Let \[I(x)=\int \frac{(x+1)}{x\left(1+x e^{x}\right)^{2}} d x, x > 0\]. If \[\lim_\limits{x \rightarrow \infty} I(x)=0\], then
I
(
1
)
I(1)
I
(
1
)
is equal to :
A
\[\frac{e+1}{e+2}-\log _{e}(e+1)\]
B
\[\frac{e+1}{e+2}+\log _{e}(e+1)\]
C
\[\frac{e+2}{e+1}-\log _{e}(e+1)\]
D
\[\frac{e+2}{e+1}+\log _{e}(e+1)\]
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Official solution
\[
I
=
∫
x
+
1
x
(
1
+
x
e
x
)
2
d
x
Put
1
+
x
e
x
=
t
⇒
x
e
x
=
t
−
1
⇒
(
x
e
x
+
e
x
)
d
x
=
d
t
⇒
e
x
(
x
+
1
)
d
x
=
d
t
\begin{aligned} & \mathrm{I}=\int \frac{x+1}{x\left(1+x e^x\right)^2} d x \\\\ & \text { Put } 1+x e^x=t \Rightarrow x e^x=t-1 \\\\ & \Rightarrow\left(x e^x+e^x\right) d x=d t \\\\ & \Rightarrow e^x(x+1) d x=d t \end{aligned}
I
=
∫
x
(
1
+
x
e
x
)
2
x
+
1
d
x
Put
1
+
x
e
x
=
t
⇒
x
e
x
=
t
−
1
⇒
(
x
e
x
+
e
x
)
d
x
=
d
t
⇒
e
x
(
x
+
1
)
d
x
=
d
t
\] \[ \therefore I=\int \frac{d t}{e^x \cdot x t^2}=\int \frac{d t}{(t-1) t^2} \] Let
1
t
2
(
t
−
1
)
=
A
(
t
−
1
)
+
B
t
+
C
t
2
\frac{1}{t^2(t-1)}=\frac{\mathrm{A}}{(t-1)}+\frac{\mathrm{B} t+\mathrm{C}}{t^2}
t
2
(
t
−
1
)
1
=
(
t
−
1
)
A
+
t
2
B
t
+
C
\[ \Rightarrow 1=\mathrm{A} t^2+(\mathrm{B} t+\mathrm{C})(t-1) \] Comparing coefficients of
t
2
,
t
t^2, t
t
2
,
t
and constant terms, we get
A
+
B
=
0
,
C
−
B
=
0
,
−
C
=
1
A+B=0, C-B=0,-C=1
A
+
B
=
0
,
C
−
B
=
0
,
−
C
=
1
On solving above equations, we get \[
C
=
−
1
,
=
B
,
A
=
1
∴
I
=
∫
1
t
−
1
d
t
+
∫
−
t
−
1
t
2
d
t
=
∫
1
t
−
1
d
t
−
∫
1
t
d
t
−
∫
1
t
2
d
t
=
log
∣
t
−
1
∣
−
log
∣
t
∣
+
1
t
+
C
⇒
I
=
log
∣
x
e
x
∣
−
log
∣
1
+
x
e
x
∣
+
1
1
+
x
e
x
+
c
\begin{aligned} & \mathrm{C}=-1,=\mathrm{B}, \mathrm{A}=1 \\\\ & \therefore \mathrm{I}=\int \frac{1}{t-1} d t+\int \frac{-t-1}{t^2} d t \\\\ & =\int \frac{1}{t-1} d t-\int \frac{1}{t} d t-\int \frac{1}{t^2} d t \\\\ & =\log |t-1|-\log |t|+\frac{1}{t}+\mathrm{C} \\\\ & \Rightarrow \mathrm{I}=\log \left|x e^x\right|-\log \left|1+x e^x\right|+\frac{1}{1+x e^x}+c \end{aligned}
C
=
−
1
,
=
B
,
A
=
1
∴
I
=
∫
t
−
1
1
d
t
+
∫
t
2
−
t
−
1
d
t
=
∫
t
−
1
1
d
t
−
∫
t
1
d
t
−
∫
t
2
1
d
t
=
lo
g
∣
t
−
1∣
−
lo
g
∣
t
∣
+
t
1
+
C
⇒
I
=
lo
g
∣
x
e
x
∣
−
lo
g
∣
1
+
x
e
x
∣
+
1
+
x
e
x
1
+
c
\] \[ =\log \left|\frac{x e^x}{1+x e^x}\right|+\frac{1}{1+x e^x}+C \] Now,
lim
x
→
∞
I
(
x
)
=
0
\lim _{x \rightarrow \infty} \mathrm{I}(x)=0
lim
x
→
∞
I
(
x
)
=
0
\[
⇒
lim
x
→
∞
{
log
∣
x
e
x
1
+
x
e
x
∣
+
1
1
+
x
e
x
+
C
}
=
0
⇒
lim
x
→
∞
{
log
∣
e
x
1
x
+
e
x
∣
+
1
x
1
x
+
e
x
+
C
}
⇒
0
+
0
+
C
=
0
⇒
C
=
0
∴
I
(
x
)
=
log
∣
x
e
x
1
+
x
e
x
∣
+
1
1
+
x
e
x
⇒
I
(
1
)
=
log
∣
e
1
+
e
∣
+
1
1
+
e
=
1
−
log
(
1
+
e
)
+
1
1
+
e
=
2
+
e
1
+
e
−
log
∣
1
+
e
∣
\begin{aligned} & \Rightarrow \lim _{x \rightarrow \infty}\left\{\log \left|\frac{x e^x}{1+x e^x}\right|+\frac{1}{1+x e^x}+\mathrm{C}\right\}=0 \\\\ & \Rightarrow \lim _{x \rightarrow \infty}\left\{\log \left|\frac{e^x}{\frac{1}{x}+e^x}\right|+\frac{\frac{1}{x}}{\frac{1}{x}+e^x}+\mathrm{C}\right\} \\\\ & \Rightarrow 0+0+\mathrm{C}=0 \Rightarrow \mathrm{C}=0 \\\\ & \therefore \mathrm{I}(x)=\log \left|\frac{x e^x}{1+x e^x}\right|+\frac{1}{1+x e^x} \\\\ & \Rightarrow \mathrm{I}(1)=\log \left|\frac{e}{1+e}\right|+\frac{1}{1+e}=1-\log (1+e)+\frac{1}{1+e} \\\\ & =\frac{2+e}{1+e}-\log |1+e| \end{aligned}
⇒
x
→
∞
lim
{
lo
g
1
+
x
e
x
x
e
x
+
1
+
x
e
x
1
+
C
}
=
0
⇒
x
→
∞
lim
{
lo
g
x
1
+
e
x
e
x
+
x
1
+
e
x
x
1
+
C
}
⇒
0
+
0
+
C
=
0
⇒
C
=
0
∴
I
(
x
)
=
lo
g
1
+
x
e
x
x
e
x
+
1
+
x
e
x
1
⇒
I
(
1
)
=
lo
g
1
+
e
e
+
1
+
e
1
=
1
−
lo
g
(
1
+
e
)
+
1
+
e
1
=
1
+
e
2
+
e
−
lo
g
∣1
+
e
∣
\]
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