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 Anti derivative of 1(x+a)(x+b)dx is equal to 

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a
logx+a+b2+x2+a+bx−ab+k
b
logx+a+b2+x2+a+bx+k
c
logx+a+b2+x2+a+bx+ab+k
d
logx−a+b2+x2+a+bx+ab+k

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detailed solution

Correct option is C

I=∫dxx+ax+bI=∫dxx2+(a+b)x+abI=∫dxx2+2(a+b)2x+ab+a+b22−a+b22I=∫dxx+a+b22+4ab−(a+b)24I=∫dxx+a+b22−a−b22I=log⁡x+a+b2+x2+(a+b)x+ab+k


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dxx2+2x+2 is equal to 


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