Question

Evaluate $\int \frac{2x^2+5x+4}{\sqrt{x^2+x+1}}dx$.

• A. $I=(x+\frac{7}{2})\sqrt{x^2+x+1}-\frac{5}{4}\log \{(x+\frac{1}{2})+\sqrt{x^2+x+1}\}+C$
• B. $I=(x+\frac{7}{2})\sqrt{x^2+x+1}+\frac{5}{4}\log \{(x+\frac{1}{2})+\sqrt{x^2+x+1}\}+C$
• C. $I=(x+\frac{3}{2})\sqrt{x^2+x+1}+\frac{1}{4}\log \{(x+\frac{1}{2})+\sqrt{x^2+x+1}\}+C$
• D. $I=(x+\frac{7}{2})\sqrt{2x^2+5x+4}+\frac{5}{4}\log \{(x+\frac{1}{2})+\sqrt{x^2+x+1}\}+C$

Answer 1

Answer: (B)

$I=(x+\frac{7}{2})\sqrt{x^2+x+1}+\frac{5}{4}\log \{(x+\frac{1}{2})+\sqrt{x^2+x+1}\}+C$

Let $I=\int \frac{2x^2+5x+4}{\sqrt{x^2+x+1}}dx$
Put $2x^2+5x+4=\lambda (x^2+x+1)+\mu \left\{\frac{d}{dx}\right(x^2+x+1\left)\right\}+\gamma$
$\Rightarrow 2x^2+5x+4=\lambda (x^2+x+1)+\mu (2x+1)+\gamma$
Comparing the coefficients of like terms, we get
$2=\lambda$, $5=\lambda +2\mu$, $4=\lambda +\mu +\gamma$
$\therefore \lambda =2$, $\mu =\frac{3}{2}$, $\gamma =\frac{1}{2}$
$\Rightarrow 2x^2+5x+4=2(x^2+x+1)+\frac{3}{2}(2x+1)+\frac{1}{2}$
Hence, the above integral reduces to
$I=\int \frac{2x^2+5x+4}{\sqrt{x^2+x+1}}dx$
$=\int (2.\frac{(x^2+x+1)}{\sqrt{x^2+x+1}}+\frac{3}{2}.\frac{(2x+1)}{\sqrt{x^2+x+1}}+\frac{1}{2}.\frac{1}{\sqrt{x^2+x+1}})dx$
$=2\int \sqrt{x^2+x+1}dx+\frac{3}{2}\int \frac{dt}{\sqrt{t}}+\frac{1}{2}\int \frac{dx}{\sqrt{x^2+x+1}}(wheret=x^2+x+1)$
$=2\int \sqrt{{(x+\frac{1}{2})}^2+{\left(\frac{\sqrt{3}}{2}\right)}^2}dx+3\sqrt{t}+\frac{1}{2}\int \frac{dx}{\sqrt{{(x+\frac{1}{2})}^2+{\left(\frac{\sqrt{3}}{2}\right)}^2}}$
$=2[\frac{1}{2}(x+\frac{1}{2})\sqrt{x^2+x+1}+\frac{1}{2}.\frac{3}{4}.\log |(x+\frac{1}{2})+\sqrt{x^2+x+1}|]+3\sqrt{x^2+x+1}+\frac{1}{2}\log |(x+\frac{1}{2})+\sqrt{x^2+x+1}|+C$
$\therefore I=(x+\frac{1}{2})\sqrt{x^2+x+1}+\frac{3}{4}\log \{(x+\frac{1}{2})+\sqrt{x^2+x+1}\}+3\sqrt{x^2+x+1}+\frac{1}{2}\log \{(x+\frac{1}{2})+\sqrt{x^2+x+1}\}+C$
$I=(x+\frac{1}{2})\sqrt{x^2+x+1}+\frac{5}{4}\log \{(x+\frac{1}{2})+\sqrt{x^2+x+1}\}+3\sqrt{x^2+x+1}+C$