questionr
Early access
Exams
Joint Entrance Examination
Mathematics
Indefinite Integrals
Question 403
Question 403 of 416
Medium
MCQ
Let \[f(x) = \int {{{2x} \over {({x^2} + 1)({x^2} + 3)}}dx} \]. If \[f(3) = {1 \over 2}({\log _e}5 - {\log _e}6)\], then
f
(
4
)
f(4)
f
(
4
)
is equal to
A
\[{\log _e}19 - {\log _e}20\]
B
\[{\log _e}17 - {\log _e}18\]
C
\[{1 \over 2}({\log _e}19 - {\log _e}17)\]
D
\[{1 \over 2}({\log _e}17 - {\log _e}19)\]
Check answer
Press A-D to select · Enter to check
Solutions · 1
Post your solution
Official solution
f
(
x
)
=
∫
2
x
(
x
2
+
1
)
(
x
2
+
3
)
d
x
f(x)=\int \frac{2 x}{\left(x^{2}+1\right)\left(x^{2}+3\right)} d x
f
(
x
)
=
∫
(
x
2
+
1
)
(
x
2
+
3
)
2
x
d
x
\[
Put
x
2
=
t
⇒
2
x
d
x
=
d
t
f
(
x
)
=
∫
d
t
(
t
+
1
)
(
t
+
3
)
=
∫
d
t
(
t
+
2
)
2
−
1
=
1
2
log
e
∣
t
+
1
t
+
3
∣
+
C
f
(
x
)
=
1
2
log
e
(
x
2
+
1
x
2
+
3
)
+
C
⇒
f
(
3
)
=
1
2
log
e
(
10
12
)
+
C
∵
f
(
3
)
+
1
2
(
log
e
5
−
log
e
6
)
⇒
C
=
0
f
(
x
)
=
1
2
log
e
(
x
2
+
1
x
2
+
3
)
⇒
f
(
4
)
=
1
2
(
log
e
17
−
log
e
19
)
\begin{aligned} & \text { Put } x^{2}=t \Rightarrow 2 x d x=d t \\\\ & f(x)=\int \frac{d t}{(t+1)(t+3)}=\int \frac{d t}{(t+2)^{2}-1} \\\\ & =\frac{1}{2} \log _{e}\left|\frac{t+1}{t+3}\right|+C \\\\ & f(x)=\frac{1}{2} \log _{e}\left(\frac{x^{2}+1}{x^{2}+3}\right)+C \Rightarrow \\\\ & f(3)=\frac{1}{2} \log _{e}\left(\frac{10}{12}\right)+C \\\\ & \because f(3)+\frac{1}{2}\left(\log _{e} 5-\log _{e} 6\right) \Rightarrow C=0 \\\\ & f(x)=\frac{1}{2} \log _{e}\left(\frac{x^{2}+1}{x^{2}+3}\right) \Rightarrow \\\\ & f(4)=\frac{1}{2}\left(\log _{e} 17-\log _{e} 19\right) \end{aligned}
Put
x
2
=
t
⇒
2
x
d
x
=
d
t
f
(
x
)
=
∫
(
t
+
1
)
(
t
+
3
)
d
t
=
∫
(
t
+
2
)
2
−
1
d
t
=
2
1
lo
g
e
t
+
3
t
+
1
+
C
f
(
x
)
=
2
1
lo
g
e
(
x
2
+
3
x
2
+
1
)
+
C
⇒
f
(
3
)
=
2
1
lo
g
e
(
12
10
)
+
C
∵
f
(
3
)
+
2
1
(
lo
g
e
5
−
lo
g
e
6
)
⇒
C
=
0
f
(
x
)
=
2
1
lo
g
e
(
x
2
+
3
x
2
+
1
)
⇒
f
(
4
)
=
2
1
(
lo
g
e
17
−
lo
g
e
19
)
\]
Answer the question to unlock
Previous
Next