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Q9  Integrate the functions ( 4x +2 ) \sqrt { x ^ 2 + x + 1 }

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Given function  ( 4x +2 ) \sqrt { x ^ 2 + x + 1 },

\int ( 4x +2 ) \sqrt { x ^ 2 + x + 1 } dx

Assume the 1+x+x^2 = t

\therefore (2x+1)dx =dt

\Rightarrow \int (4x+2)\sqrt{1+x+x^2} dx

= \int 2\sqrt {t}dt = 2\int \sqrt{t}dt

= 2\left ( \frac{t^{\frac{3}{2}}}{\frac{3}{2}} \right ) +C

Now, back substituting the value of t in the above equation,

 = \frac{4}{3}(1+x+x^2)^{\frac{3}{2}} +C, where C is any constant value.

Posted by

Divya Prakash Singh

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