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Question

Evaluate 1(1x2)1+x2 dx


A
122log1+x22x1+x2+2x+c
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B
12log1+x21+x2+2x+c
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C
1+x22x1+x2+2x+c
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D
152log1+x22x1+x2+c
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Solution

The correct option is A 122log1+x22x1+x2+2x+c
Putting x=1t and dx=1t2dt, we get
I=(1t2)dt(11t2)1+1t2

I=(1t2)×t2×t dt(1t2)1+t2


I=t dt(t21)t2+1

Let t2+1=u2 and t2+1=u

2t dt=2u du

t dt=u du

Since, t2+1=u2

t2+12=u22

t21=u22

Now,

I=t dt(t21)t2+1

I=u du(u22)u

I=du(u22)
I=duu2(2)2

=122logu2u+2+c [Since, dxx2a2=12alogxax+a+c]


=122logt2+12t2+1+2+c


=122log∣ ∣ ∣ ∣1x2+121x2+1+2∣ ∣ ∣ ∣+c


=122log∣ ∣ ∣ ∣1+x2x21+x2x+2∣ ∣ ∣ ∣+c


=122log1+x22x1+x2+2x+c


1(1x2)1+x2 dx=122log1+x22x1+x2+2x+c


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