I=∫x2+x+1dx is equal to

I=∫x2+x+1dx=∫x+122−14+1dx=∫x+122+322dx Use :∫x2+a2=x2x2+a2+a22log⁡x+x2+a2I=x+122x2+x+1+38logx+12x2+x+1+C

I=∫x2+x+1dx is equal to

I=∫x2+x+1dx=∫x+122−14+1dx=∫x+122+322dx Use :∫x2+a2=x2x2+a2+a22log⁡x+x2+a2I=x+122x2+x+1+38logx+12x2+x+1+C

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$I = \int \sqrt{x^{2} + x + 1} d x is equal to$

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$\frac{(x + \frac{1}{2})}{2} \sqrt{x^{2} + x + 1} + \frac{3}{8} \log |x + \frac{1}{2} + \sqrt{x^{2} + x + 1}| + C$

$\frac{x + \frac{1}{2}}{2} \sqrt{x^{2} + x + 1} - \frac{3}{8} \log |x + \frac{1}{2} + \sqrt{x^{2} + x + 1}| + C$

$\frac{x + \frac{1}{2}}{2} \sqrt{x^{2} + x + 1} + \frac{3}{4} \log |x + \frac{1}{2} + \sqrt{x^{2} + x + 1}| + C$

$\frac{x + \frac{1}{2}}{2} \sqrt{x^{2} + x + 1} - \frac{3}{4} \log |x + \frac{1}{2} + \sqrt{x^{2} + x + 1}| + C$

answer is C.

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Detailed Solution

$I = \int \sqrt{x^{2} + x + 1} d x = \int \sqrt{{(x + \frac{1}{2})}^{2} - \frac{1}{4} + 1} d x$

$= \int \sqrt{{(x + \frac{1}{2})}^{2} + {(\frac{\sqrt{3}}{2})}^{2}} d x$

$Use : \int \sqrt{x^{2} + a^{2}} = \frac{x}{2} \sqrt{x^{2} + a^{2}} + \frac{a^{2}}{2} \log |x + \sqrt{x^{2} + a^{2}}|$

$I = \frac{(x + \frac{1}{2})}{2} \sqrt{x^{2} + x + 1} + \frac{3}{8} \log |x + \frac{1}{2} \sqrt{x^{2} + x + 1}| + C$

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$I = \int \sqrt{x^{2} + x + 1} d x is equal to$