Question

Evaluate : $$\int (3x-2)\sqrt{x^2+x+1}dx$$ .

Answer 1

$$I=\int (3x-2)\sqrt{x^2+x+1}dx$$

$$\text{ Let }3x-2=a(2x+1)+b$$

$$\Rightarrow a=\frac{3}{2}\text{ and }b=\frac{-7}{2}$$

$$\text{ So,}I=\frac{3}{2}\int (2x+1)\sqrt{x^2+x+1}dx-\frac{7}{2}\int \sqrt{x^2+x+1}dx$$

$$\text{ Let }I=\frac{3}{2}{I}_1-\frac{7}{2}{I}_2\dots \left(1\right)$$

$$\text{ Here },I_1=\int (2x+1)\sqrt{x^2+x+1}dx\text{ and }{I}_2=\int \sqrt{x^2+x+1}dx$$

$$\text{ Now },I_1=\int (2x+1)\sqrt{x^2+x+1}dx$$

$$\text{ Let }{x}^2+x+1=t$$

$$\Rightarrow (2x+1)dx=dt$$

$$\text{ So },I_1=\int \sqrt{t}dt=\frac{2}{3}{t}^{\frac{3}{2}}=\frac{2}{3}(x^2+x+1{)}^{\frac{3}{2}}+c_1\dots \left(2\right)$$

$$\text{ And }{I}_2=\int \sqrt{x^2+x+1}dx=\int \sqrt{{(x+\frac{1}{2})}^2+{\left(\frac{\sqrt{3}}{2}\right)}^2}dx$$

$$=\frac{x+\frac{1}{2}}{2}\sqrt{x^2+x+1}+\frac{3}{8}\log |x+\frac{1}{2}+\sqrt{x^2+x+1}|+c_2\dots \left(3\right)$$

By putting the values of equation (2) and equation (3) in equation (1), we get:

$$I={(x^2+x+1)}^{\frac{3}{2}}-\frac{7}{2}[\left(\frac{2x+1}{4}\right)\sqrt{x^2+x+1}+\frac{3}{8}\log |(x+\frac{1}{2})+\sqrt{x^2+x+1}|]+c$$