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Question

Integrate
(4x+2)x2+x+1

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Solution

(4x+2)x2+x+1dx

Let t=x2+x+1dt=(2x+1)dx

=2(2x+1)x2+x+1dx

=2tdt

=2t12dt

=2⎢ ⎢ ⎢t12+112+1⎥ ⎥ ⎥+c

=2×23t32+c where t=x2+x+1

=43(x2+x+1)32+c

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