Mathematics

Evaluate the following integral:
$$\int { \left( x+2 \right)  } \sqrt { { x }^{ 2 }+x+1 } dx$$


SOLUTION
Let $$I=\displaystyle\int (x+2)\sqrt{x^2+x+1}\ d x$$

         $$=\displaystyle\int ( x+2)\sqrt{x^2+2(x)\left(\dfrac{1}{2}\right)+1}\ d x$$

         $$=\displaystyle\int (x+2)\sqrt{x^2+2(x)\left(\dfrac{1}{2}\right)+\bigg(\dfrac{1}{2}\bigg)^2-\bigg(\dfrac{1}{2}\bigg)^2+1} \ \ d x$$

         $$=\displaystyle\int (x+2)\sqrt{\bigg(x+\dfrac{1}{2}\bigg)^2+\dfrac{3}{4}}\ d x$$                       $$(\because (a+b)^2=a^2+2 a b+b^2)$$

Put $$x+\dfrac{1}{2}=t\implies x=t-\dfrac{1}{2}\implies d x=d t$$

$$\implies I=\displaystyle\int \bigg(t-\dfrac{1}{2}+2\bigg)\sqrt{t^2+\dfrac{3}{4}}\ d t$$

$$\implies I=\displaystyle\int \bigg( t+\dfrac{3}{2}\bigg)\sqrt{t^2+\dfrac{3}{4}}\ d t$$

$$\implies I=\displaystyle\int t\sqrt{t^2+\dfrac{3}{4}}\ d t+\dfrac{3}{2}\int \sqrt{t^2+\left(\dfrac{\sqrt{3}}{2}\right)^2}\  d t$$

As we know that

$$\displaystyle\int \sqrt{x^2+a^2}\ d x=\dfrac{x}{2}\sqrt{x^2+a^2}+\dfrac{a^2}{2}\ln |x+\sqrt{x^2+a^2}|+c$$

$$\implies I=\displaystyle\int t\sqrt{t^2+\dfrac{3}{4}}\ d t+\dfrac{3}{2}\left(\dfrac{t}{2}\sqrt{t^2+\dfrac{3}{4}}+\dfrac{3/4}{2}\ln \left|t+\sqrt{t^2+\dfrac{3}{4}}\right|+c\right)$$

Put $$t^2+\dfrac{3}{4}=y\implies 2 t d t=d y\implies t d t=\dfrac{dy}{2}$$

$$\implies I=\displaystyle\int \sqrt{y} \times \dfrac{d y}{2}+\dfrac{3 t}{4}\sqrt{t^2+\dfrac{3}{4}}+\dfrac{9}{16}\ln \left|t+\sqrt{t^2+\dfrac{3}{4}}\right|+c$$

$$\implies I=\dfrac{1}{2}\displaystyle\int y^{1/2}\ d y+\dfrac{3 t}{4}\sqrt{t^2+\dfrac{3}{4}}+\dfrac{9}{16}\ln \left|t+\sqrt{t^2+\dfrac{3}{4}}\right|+c$$

$$\implies I=\dfrac{1}{2}\times \dfrac{y^{3/2}}{3/2}+\dfrac{3 t}{4}\sqrt{t^2+\dfrac{3}{4}}+\dfrac{9}{16}\ln \left|t+\sqrt{t^2+\dfrac{3}{4}}\right|+C$$                                       $$\left(\because \displaystyle\int x^n\ d x=\dfrac{x^{n+1}}{n+1}+c\right)$$

$$\implies I=\dfrac{1}{3} y^{3/2}+\dfrac{3 t}{4}\sqrt{t^2+\dfrac{3}{4}}+\dfrac{9}{16}\ln \left|t+\sqrt{t^2+\dfrac{3}{4}}\right|+C$$

$$\implies I=\dfrac{1}{3} \left(t^2+\dfrac{3}{4}\right)^{3/2}+\dfrac{3 t}{4}\sqrt{t^2+\dfrac{3}{4}}+\dfrac{9}{16}\ln \left|t+\sqrt{t^2+\dfrac{3}{4}}\right|+C$$                      $$\bigg(\because y=t^2+\dfrac{3}{4}\bigg)$$

$$\implies I=\dfrac{1}{3} \left(\bigg(x+\dfrac{1}{2}\bigg)^2+\dfrac{3}{4}\right)^{3/2}+\dfrac{3(x+1/2)}{4}\sqrt{\left(x+\dfrac{1}{2}\right)^2+\dfrac{3}{4}}+\dfrac{9}{16}\ln \left|x+\dfrac{1}{2}+\sqrt{\bigg(x+\dfrac{1}{2}\bigg)^2+\dfrac{3}{4}}\right|+C$$

                                                                                                                                                                     $$\bigg(\because t=x+\dfrac{1}{2}\bigg)$$

$$\implies I=\dfrac{1}{3} (x^2+x+1)^{3/2}+\dfrac{3}{8}(2 x+1)\sqrt{x^2+x+1}+\dfrac{9}{16}\ln \left|x+\dfrac{1}{2}+\sqrt{x^2+x+1}\right|+C$$   
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