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Joint Entrance Examination
Mathematics
Indefinite Integrals
Question 317
Question 317 of 416
Medium
MCQ
If
∫
x
2
−
x
+
1
x
2
+
1
e
cot
−
1
x
d
x
=
A
(
x
)
e
cot
−
1
x
+
C
\displaystyle\int \dfrac{x^2-x+1}{x^2+1}e^{\cot^{-1}x}\,dx = A(x)e^{\cot^{-1}x}+C
∫
x
2
+
1
x
2
−
x
+
1
e
c
o
t
−
1
x
d
x
=
A
(
x
)
e
c
o
t
−
1
x
+
C
, then
A
(
x
)
A(x)
A
(
x
)
is equal to:
A
−
x
-x
−
x
B
x
x
x
C
1
−
x
\sqrt{1-x}
1
−
x
D
1
+
x
\sqrt{1+x}
1
+
x
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Solutions · 1
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Official solution
Put
x
=
cot
t
⇒
−
cosec
2
t
d
t
=
d
x
x=\cot t \Rightarrow -\cosec^2t\,dt = dx
x
=
cot
t
⇒
−
cosec
2
t
d
t
=
d
x
; also
1
+
cot
2
t
=
cosec
2
t
1+\cot^2t = \cosec^2t
1
+
cot
2
t
=
cosec
2
t
I
=
∫
e
t
(
cot
2
t
−
cot
t
+
1
)
1
+
cot
2
t
(
−
cosec
2
t
)
d
t
=
−
∫
e
t
(
cosec
2
t
−
cot
t
)
d
t
I = \int \dfrac{e^t(\cot^2t-\cot t+1)}{1+\cot^2t}(-\cosec^2t)\,dt = -\int e^t(\cosec^2t-\cot t)\,dt
I
=
∫
1
+
cot
2
t
e
t
(
cot
2
t
−
cot
t
+
1
)
(
−
cosec
2
t
)
d
t
=
−
∫
e
t
(
cosec
2
t
−
cot
t
)
d
t
=
∫
e
t
(
cot
t
−
cosec
2
t
)
d
t
=
e
t
cot
t
+
C
=
e
cot
−
1
x
⋅
x
+
C
= \int e^t(\cot t - \cosec^2t)\,dt = e^t\cot t + C = e^{\cot^{-1}x}\cdot x+C
=
∫
e
t
(
cot
t
−
cosec
2
t
)
d
t
=
e
t
cot
t
+
C
=
e
c
o
t
−
1
x
⋅
x
+
C
∴
A
(
x
)
=
x
\therefore A(x) = x
∴
A
(
x
)
=
x
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