Question

integration of (3x-2)√ x²+x+1 dx

Answer 1

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Answer 2

EXPLANATION:

Now, $\displaystyle \mathrm{\int (3x-2)\sqrt{x^{2}+x+1}\:dx}$

$\displaystyle \mathrm{=\frac{3}{2} \int (2x+1-\frac{4}{3})\sqrt{x^{2}+x+1}\:dx}$

$\displaystyle \mathrm{=\frac{3}{2} \int (2x+1)\sqrt{x^{2}+x+1}\:dx-2\int \sqrt{x^{2}+x+1}\:dx}$

$\displaystyle \mathrm{=\frac{3}{2} \int (x^{2}+x+1)^{\frac{1}{2}}\:d(x^{2}+x+1)-2\int \sqrt{(x+\frac{1}{2})^{2}-(\frac{\sqrt{3}}{2})^{2}}\:dx}$

$\displaystyle \mathrm{=\frac{3}{2}\frac{(x^{2}+x+1)^{\frac{3}{2}}}{\frac{3}{2}}-2[\frac{x+\frac{1}{2}}{2}-\frac{(\frac{\sqrt{3}}{2})^{2}}{2}ln|x+\frac{1}{2}+\sqrt{x+\frac{1}{2}-(\frac{\sqrt{3}}{2})^{2}}|]+C}$

where C is any arbitrary constant

$\displaystyle \mathrm{=(x^{2}+x+1)^{\frac{3}{2}}-(x+\frac{1}{2})\sqrt{x^{2}+x+1}+\frac{3}{4}ln|x+\frac{1}{2}+\sqrt{x^{2}+x+1}|+C}$

This is the required integral.

FORMULA:

$\displaystyle\mathrm{\int \sqrt{x^{2}-a^{2}}\:dx=\frac{x}{2}\sqrt{x^{2}-a^{2}}-\frac{a^{2}}{2}ln|x+\sqrt{x^{2}-a^{2}}|+C}$

where C is integral constant.