Mathematics

Evaluate the following integral:
$$\int { \left( x+1 \right)  } \sqrt { 2{ x }^{ 2 }+3 } dx$$


SOLUTION
Let $$I=\displaystyle\int (x+1)\sqrt{2 x^2+3}\ d x$$

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

$$\implies I=\sqrt{2}\displaystyle\int (x+1)\sqrt{x^2+\dfrac{3}{2}}\ d x$$

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

As we know that

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

$$\implies I=\sqrt{2}\displaystyle\int x\sqrt{x^2+\dfrac{3}{2}}\ d x+\sqrt{2}\left(\dfrac{x}{2}\sqrt{x^2+\dfrac{3}{2}}+\dfrac{3/2}{2}\ln \bigg| x+\sqrt{x^2+\dfrac{3}{2}}\ \bigg|\right)+C$$

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

$$\implies I=\sqrt{2}\displaystyle\int \sqrt{y}\times \dfrac{d y}{2}+\dfrac{x}{\sqrt{2}}\sqrt{x^2+\dfrac{3}{2}}+\dfrac{3}{2\sqrt{2}}\ln \bigg| x+\sqrt{x^2+\dfrac{3}{2}}\ \bigg|+C$$

$$\implies I=\dfrac{1}{\sqrt{2}}\displaystyle\int \sqrt{y}\ d y+\dfrac{x}{\sqrt{2}}\sqrt{x^2+\dfrac{3}{2}}+\dfrac{3}{2\sqrt{2}}\ln \bigg| x+\sqrt{x^2+\dfrac{3}{2}}\ \bigg|+C$$  

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


$$\implies I=\dfrac{\sqrt{2}}{3}\ y^{3/2}+\dfrac{\sqrt{2}x}{2}\sqrt{x^2+\dfrac{3}{2}}+\dfrac{3}{2\sqrt{2}}\ln \bigg| x+\sqrt{x^2+\dfrac{3}{2}}\ \bigg|+C$$   

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

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

$$\implies I=\dfrac{1}{6}\ \left(2 x^2+3\right)^{3/2}+\dfrac{x}{2}\sqrt{2 x^2+3}+\dfrac{3}{2\sqrt{2}}\ln \bigg| x+\sqrt{x^2+\dfrac{3}{2}}\ \bigg|+C$$      
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