∫x(x-1)(x2+4)dx

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15logx-1-110log(x2+4)-25tan-1x2+C

15logx-1-110log(x2+4)+45tan-1x2+C

15logx-1-110log(x2+4)+25tan-1x2+C

None of these

answer is C.

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Detailed Solution

∫x(x-1)(x2+4)dx let x(x-1)(x2+4)=Ax-1+Bx+Cx2+4 --------(1) or x=A(x2+4)+(Bx+C)(x-1) -------(2) putting x=1 in equation (2), we get 1=5A or A=15 putting x=0 in equation (2), we get 0=4A-C or C=4A=45 putting x=-1 in equation (2), we get -1=5A+2B-2C solving these equations, we get A=15, B=-15 and C=45 Substituting the values of A,B and C in equation (1) ,we obtain x(x-1)(x2+4)=15(x-1)+-15x+45x2+4 =15(x-1)-15(x-4)(x2+4) I =15∫1(x-1)dx-15∫x-4x2+4 dx =15∫1(x-1)dx-110∫2xx2+4 dx+45∫1x2+4 dx =15logx-1-110log(x2+4)+45×12tan-1x2+C 15logx-1-110log(x2+4)+25tan-1x2+C

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