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a
x+122+x2+x+1+38logx+12+x2+x+1+3x2+x+1+92logx+12+x2+x+1+C
b
x+12+x2+x+1+18logx+12+x2+x+1+x2+x+1+12logx+12+x2+x+1+C
c
x−12+x2+x+1+18logx+1−+x2+x+1−x2+x+1+12logx+−2+x2+x+1+C
d
None of the above
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Correct option is A
Let x2+4x+7=Ax2+x+1+B(2x+1)+CComparing the coefficients of x2, x and constant term, we getA=1,A+2B=4,A+B+C=7⇒A=1,B=32,C=92So, I=∫x2+x+1dx+32∫(2x+1)x2+x+1dx+92∫dxx2+x+1Now,x2+x+1=x+122+322I=x+122⋅x2+x+1+38logx+12+x2+x+1+3x2+x+1+92logx+12+x2+x+1+CSimilar Questions
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