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 Evaluate (1+x)3xdx

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a
2x+2x3/2+65x5/2+27x7/2+c
b
2x+x3/2+65x5/2+27x7/2+c
c
2x+2x3/2+25x5/2+27x7/2+c
d
2x+23x3/2+65x5/2+27x7/2+c

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detailed solution

Correct option is A

∫(1+x)3xdx=∫1+3x+3x2+x3xdx=∫x−1/2dx+3∫x1/2dx+3∫x3/2dx+∫x5/2dx=x1/21/2+3x3/23/2+3⋅x5/25/2+x7/27/2+c=2x+2x3/2+65x5/2+27x7/2+c


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If x+1x2+x+1dxx2+x+1=kx1x2+x+1+C then the value of k


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