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Question

The integral of dxx13(x61) is:

A
16lnx61x6+c
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B
16[lnx61x6+x6+12x12]+c
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C
16[lnx61x6+x6+x12]+c
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D
16[lnx61x6+x6+14x12]+c
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Solution

The correct option is B 16[lnx61x6+x6+12x12]+c
I=dxx13(x61)=x5dxx18(x61)
Put x6=t6x5dx=dt
I=dt6t3(t1)=16dtt3(t1)
Using partial fractions, we have:
1t3(t1)=At1+Bt2+Ct+Dt3
1t3(t1)=At3+Bt3+Ct2Bt2+DtCtDt3(t1)
On comparing the coefficents, we have:
D=C=B=1;A=1
I=16(1t1t2+t+1t3)dt
I=16(1t11t1t21t3)dt
I=16[ln(t1t)t11t22]+c
I=16[lnx61x6+x6+12x12]+c

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