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# Integrate : $\int \large\frac{3x+1}{(3x^2+2x+1)^2}$$dx $\begin {array} {1 1} (a)\;\frac{1}{2}(3x^2+2x+1)+c \\ (b)\;(3x^2+2x+1)+c \\ (c)\;-\frac{1}{2}(3x^2+2x+1) +c \\ (d)\;None \end {array}$ ## 1 Answer Comment A) \int \large\frac{3x+1}{(3x^2+2x+1)^2}$$dx$
=> $3x^2 +2x+1=t$
=> $(6x+2) dx=dt$
=> $(3x+1)dx =\large\frac{1}{2}.$$dt \large\frac{1}{2} \int \large\frac{dt}{t^2} => \large\frac{-1}{2}$$t+c$
=> $\large\frac{-1}{2}$$(3x^2+2x+1)+c$
Hence c is the correct answer.
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