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Joint Entrance Examination
Mathematics
Indefinite Integrals
Question 316
Question 316 of 416
Medium
MCQ
The integral
∫
x
d
x
2
−
x
2
+
2
−
x
2
\displaystyle\int \dfrac{x\,dx}{\sqrt{2-x^2}+2-x^2}
∫
2
−
x
2
+
2
−
x
2
x
d
x
equals:
A
log
∣
1
+
2
+
x
2
∣
+
c
\log\left|1+\sqrt{2+x^2}\right|+c
lo
g
1
+
2
+
x
2
+
c
B
−
log
∣
1
+
2
−
x
2
∣
+
c
-\log\left|1+\sqrt{2-x^2}\right|+c
−
lo
g
1
+
2
−
x
2
+
c
C
x
log
∣
1
−
2
+
x
2
∣
+
c
x\log\left|1-\sqrt{2+x^2}\right|+c
x
lo
g
1
−
2
+
x
2
+
c
D
−
x
log
∣
1
−
2
−
x
2
∣
+
c
-x\log\left|1-\sqrt{2-x^2}\right|+c
−
x
lo
g
1
−
2
−
x
2
+
c
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Solutions · 1
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Official solution
I
=
∫
x
d
x
2
−
x
2
+
(
2
−
x
2
)
I = \int \dfrac{x\,dx}{\sqrt{2-x^2}+(2-x^2)}
I
=
∫
2
−
x
2
+
(
2
−
x
2
)
x
d
x
Put
t
=
2
−
x
2
t = \sqrt{2-x^2}
t
=
2
−
x
2
,
d
t
d
x
=
−
x
2
−
x
2
⇒
−
t
d
t
=
x
d
x
\dfrac{dt}{dx} = \dfrac{-x}{\sqrt{2-x^2}} \Rightarrow -t\,dt = x\,dx
d
x
d
t
=
2
−
x
2
−
x
⇒
−
t
d
t
=
x
d
x
I
=
∫
−
t
d
t
t
2
+
t
=
−
∫
d
t
t
+
1
=
−
log
∣
t
+
1
∣
I = \int \dfrac{-t\,dt}{t^2+t} = -\int \dfrac{dt}{t+1} = -\log|t+1|
I
=
∫
t
2
+
t
−
t
d
t
=
−
∫
t
+
1
d
t
=
−
lo
g
∣
t
+
1∣
=
−
log
∣
1
+
2
−
x
2
∣
+
c
= -\log\left|1+\sqrt{2-x^2}\right|+c
=
−
lo
g
1
+
2
−
x
2
+
c
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