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a
x+22x2+4x+6+log(x+2)+x2+4x+6+C
b
(x+2)x2+4x+6+12log(x+2)+x2+4x+6+C
c
(x+2)x2+4x+6−12log(x+2)+x2+4x+6+C
d
None of the above
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Correct option is A
Here, we can convert the integrand into some standard integrand of the form ,x2±a2 and then integrate. Let I=∫x2+4x+6dx =∫x2+4x+22+6−4dx=∫(x+2)2+(2)2dx∵∫x2+a2dx=x2x2+a2+a22log∣x+x2+a2 =x+22x2+4x+6+22log(x+2)+x2+4x+6+C =x+22x2+4x+6+log(x+2)+x2+4x+6+CSimilar Questions
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