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Question

[x+a2+x2]na2+x2 dx (n0)=

A
1n+1(x+a2+x2)n+1+c
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B
1n.(x+a2+x2)n+c
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C
(x+a2+x2)n+c
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D
1n(x+a2+x2)n+c
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Solution

The correct option is D 1n(x+a2+x2)n+c
Put x+a2+x2=tthen [1+2x2a2+x2] dx=dt[x+a2+x2]na2+x2 dx=tn1 dt=tnn+c=1n(x+a2+x2)n+c

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