Question

13)evaluate (x+1)√2x^2+3 ​

Answer 1

$$\begin{gathered} \mathrm{Let}\mathrm{I}=\int \left(\mathrm{x}+1\right)\sqrt{2{\mathrm{x}}^2+3}\mathrm{dx} \\ =\int \mathrm{x}\sqrt{2{\mathrm{x}}^2+3}\mathrm{dx}+\int \sqrt{2{\mathrm{x}}^2+3}\mathrm{dx} \\ \mathrm{Let}{\mathrm{I}}_1=\int \mathrm{x}\sqrt{2{\mathrm{x}}^2+3}\mathrm{dx} \\ \mathrm{put}2{\mathrm{x}}^2+3=\mathrm{t} \\ \Rightarrow 4\mathrm{x}\mathrm{dx}=\mathrm{dt} \\ \Rightarrow \mathrm{x}\mathrm{dx}=\frac{\mathrm{dt}}{4} \\ \mathrm{Now},{\mathrm{I}}_1=\frac{1}{4}\int \sqrt{\mathrm{t}}\mathrm{dt} \\ =\frac{1}{4}\times \frac{2}{3}\times {\mathrm{t}}^{3/2} \\ =\frac{1}{6}{\left(2{\mathrm{x}}^2+3\right)}^{3/2} \\ \mathrm{Let}{\mathrm{I}}_2=\int \sqrt{2{\mathrm{x}}^2+3}\mathrm{dx} \\ =\sqrt{2}\int \sqrt{{\mathrm{x}}^2+\frac{3}{2}}\mathrm{dx} \\ =\sqrt{2}\int \sqrt{{\mathrm{x}}^2+{\left(\frac{\sqrt{3}}{\sqrt{2}}\right)}^2}\mathrm{dx} \\ =\sqrt{2}\times \left[\frac{\mathrm{x}}{2}\sqrt{{\mathrm{x}}^2+\frac{3}{2}}+\frac{3}{4}\log \left|\mathrm{x}+\sqrt{{\mathrm{x}}^2+\frac{3}{2}}\right|\right] \\ \mathrm{So},\mathrm{I}=\frac{1}{6}{\left(2{\mathrm{x}}^2+3\right)}^{3/2}+\sqrt{2}\times \left[\frac{\mathrm{x}}{2}\sqrt{{\mathrm{x}}^2+\frac{3}{2}}+\frac{3}{4}\log \left|\mathrm{x}+\sqrt{{\mathrm{x}}^2+\frac{3}{2}}\right|\right]+\mathrm{C} \end{gathered}$$