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Question

If pR and C is arbitrary constant of integration, then (ex+1)dxe2x+x2+2xexp31 is equal to

A
1p31tan1(ex+xp31)+C for p<1
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B
12p31ln∣ ∣ex+xp31ex+x+p31∣ ∣+C for p<1
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C
1ex+x+C for p=1
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D
12p3+1ln∣ ∣ex+xp3+1ex+x+p3+1∣ ∣+C for p>1
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Solution

The correct option is D 12p3+1ln∣ ∣ex+xp3+1ex+x+p3+1∣ ∣+C for p>1
I=(ex+1)dxe2x+x2+2xexp31
=(ex+1)dx(ex+x)2p31
Put ex+x=t(ex+1)dx=dt
I=dtt2p31

For p=1,
I=dtt2
=1t+C
=1ex+x+C


For p<1,
I=dtt2+(p31)+ve
=1p31tan1(ex+xp31)+C


For p>1,
I=dtt2(p3+1)+ve
=12p3+1ln∣ ∣ex+xp3+1ex+x+p3+1∣ ∣+C

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