Question

Solve

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

Answer 1

Let $(3x-2)=A\times \frac{d}{dx}(x^2+x+1)+B$

$\Rightarrow (3x-2)=A\times (x^2+1)+B$

$\Rightarrow 3x-2=2Ax+A+B$

Comparing coefficient of both sides

$2A=3\Rightarrow A=\frac{3}{2}$

and $A+B=-2$

so, $B=-2-\frac{3}{2}$

$B=-\frac{7}{2}$

Now, $3x-2=\frac{3}{2}(2x+1)-\frac{7}{2}$

Now, $I=\int \left[\frac{3}{2}\right(2x+1)-\frac{7}{2}]\sqrt{x^2+x+1}dx$

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

$I=\frac{3}{2}{I}_1-\frac{7}{2}{I}_2$ __(A)

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

Put $x^2+x+1=t$

$(2x+1)dx=dt$

Now, $I_1=\int \sqrt{t}dt=\frac{2}{3}{t}^{3/2}+C_1$

$I_1=\frac{2}{3}(x^2+x+1{)}^{3/2}+C_1$ __(B)

Now, $I_2=\int \sqrt{x^2+x+1}dx$

$=\int \sqrt{x^2+x+\frac{1}{4}-\frac{1}{4}+1}dx$

$I_2=\int \sqrt{{(x+\frac{1}{2})}^2+{\left(\frac{\sqrt{3}}{2}\right)}^2}dx$

$=\frac{1}{2}(x+\frac{1}{2})\sqrt{(x+\frac{1}{2}{)}^2-(\frac{\sqrt{3}}{2}{)}^2}+\frac{1}{2}\times \frac{3}{4}\log /(x+\frac{1}{2})+$

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

from eq (A) , (B) and (C)

$I=(x^2+x+1{)}^{3/2}-\frac{7}{4}[(x+\frac{1}{2})\sqrt{x^2+x+1}+\frac{3}{4}log$

$I=(x^2+x+1{)}^{3/2}-\frac{7(2x+1)}{8}\sqrt{x^2+x+1}+\frac{21}{16}\log |x+\frac{1}{2}+$

where, $C=C_1+C_2$