Question

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

• A. $\frac{2}{3}(x^2+x+1{)}^{\frac{3}{2}}+C$
• B. $\frac{2}{5}(x^2+x+1{)}^{\frac{5}{2}}+C$
• C. $(x^2+x+1{)}^{\frac{3}{2}}+C$
• D. $\frac{2}{3}(x^2+x+1{)}^{\frac{1}{2}}+C$

Answer 1

The correct option is A $\frac{2}{3}(x^2+x+1{)}^{\frac{3}{2}}+C$

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

$=\int f^{\prime }\left(x\right)\sqrt{f\left(x\right)}dx$

$=\int f^{\prime }\left(x\right)f^n\left(x\right)dx$

$=\frac{f^{n+1}\left(x\right)}{n+1}$

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

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

$=\frac{(x^2+x+1{)}^{\frac{1}{2}+1}}{1+\frac{1}{2}}+C=\frac{2}{3}(x^2+x+1{)}^{\frac{3}{2}}+C$.