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a
123loglogx+2−3logx+2+3+c
b
123loglogx+2+3logx+2−3+c
c
123log2+3+logx2+3−logx+c
d
123log2+3−logx2+3+logx+c
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Correct option is A
∫dxxlogx2+4logx+1 Put logx=t1xdx=dt∫dxx(logx)2+4logx+1=∫dtt2+4t+1 Use ∫1x2−a2dx=12alogx−ax+a+c=∫dtt2+2.2t+(2)2+1−(2)2=∫dt(t+2)2−(3)=∫dt(t+2)2−(3)2=123logt+2−3t+2+3+c=123loglogx+2−3logx+2+3Similar Questions
is equal to
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