Question

integrate;
1)dx/ xlogx log(logx)
2)dx/ (x²-1) under root x²+1
3)x⁵dx/under root 1+x³
4)x under root 1+x² dx
5)x+1 dx / under root x²+1

Answer 1

$$\begin{gathered} \left(1\right): \\ \mathrm{I}=\int \frac{\mathrm{dx}}{\mathrm{x}\mathrm{logx}\log \left(\mathrm{logx}\right)} \\ \mathrm{Let},\log \left(\mathrm{logx}\right)=\mathrm{t} \\ \Rightarrow \frac{1}{\mathrm{x}\mathrm{logx}}\mathrm{dx}=\mathrm{dt}\left[\mathrm{As},\frac{\mathrm{d}}{\mathrm{dx}}\log \left\{\mathrm{f}\left(\mathrm{x}\right)\right\}=\frac{1}{\mathrm{f}\left(\mathrm{x}\right)}\times \mathrm{f}\text{'}\left(\mathrm{x}\right)\mathrm{and}\frac{\mathrm{d}}{\mathrm{dx}}\left(\mathrm{logx}\right)=\frac{1}{\mathrm{x}}\right] \\ \mathrm{Therefore}, \\ \mathrm{I}=\int \frac{1}{\mathrm{t}}\mathrm{dt}=\log \left(\mathrm{t}\right)+\mathrm{C}\left[\mathrm{As},\int \frac{1}{\mathrm{x}}\mathrm{dx}=\log \left(\mathrm{x}\right)+\mathrm{C}\right] \\ \Rightarrow \mathrm{I}=\log \left\{\log \left(\mathrm{logx}\right)\right\}+\mathrm{C} \\ \left(2\right) \\ \mathrm{I}=\int \frac{1}{\left({\mathrm{x}}^2-1\right)\sqrt{{\mathrm{x}}^2+1}}\mathrm{dx} \\ \mathrm{Put},\mathrm{x}=\frac{1}{\mathrm{t}}\Rightarrow \mathrm{dx}=-\frac{1}{{\mathrm{t}}^2}\mathrm{dt} \\ \mathrm{Therefore}, \\ \mathrm{I}=\int \frac{\left(-{\displaystyle \frac{1}{{\mathrm{t}}^2}}\right)}{\left({\displaystyle \frac{1}{{\mathrm{t}}^2}}-1\right)\sqrt{{\displaystyle \frac{1}{{\mathrm{t}}^2}}+1}}\mathrm{dt}=-\int \frac{\mathrm{t}\mathrm{dt}}{\left(1-{\mathrm{t}}^2\right)\sqrt{1+{\mathrm{t}}^2}} \\ \mathrm{Now},\mathrm{put}1+{\mathrm{t}}^2={\mathrm{u}}^2\Rightarrow 2\mathrm{t}\mathrm{dt}=2\mathrm{u}\mathrm{du}\Rightarrow \mathrm{t}\mathrm{dt}=\mathrm{u}\mathrm{du} \\ \mathrm{Therefore}, \\ \mathrm{I}=-\int \frac{\mathrm{u}\mathrm{du}}{\left(2-{\mathrm{u}}^2\right)\sqrt{{\mathrm{u}}^2}}=-\int \frac{\mathrm{u}\mathrm{du}}{-\left({\mathrm{u}}^2-2\right)\mathrm{u}}=\int \frac{\mathrm{du}}{{\mathrm{u}}^2-{\left(\sqrt{2}\right)}^2} \\ \Rightarrow \mathrm{I}=\int \frac{1}{\left(\mathrm{u}-\sqrt{2}\right)\left(\mathrm{u}+\sqrt{2}\right)}\mathrm{du}=\frac{1}{2\sqrt{2}}\int \left(\frac{1}{\mathrm{u}-\sqrt{2}}-\frac{1}{\mathrm{u}+\sqrt{2}}\right)\mathrm{du} \\ \Rightarrow \mathrm{I}=\frac{1}{2\sqrt{2}}\left(\log \left|\mathrm{u}-\sqrt{2}\right|-\log \left|\mathrm{u}+\sqrt{2}\right|\right)+\mathrm{C}\left(\mathrm{As},\int \frac{1}{\mathrm{x}+\mathrm{a}}\mathrm{dx}=\log \left|\mathrm{x}+\mathrm{a}\right|+\mathrm{C}\right) \\ \Rightarrow \mathrm{I}=\frac{1}{2\sqrt{2}}\log \left|\frac{\mathrm{u}-\sqrt{2}}{\mathrm{u}+\sqrt{2}}\right|+\mathrm{C} \\ \Rightarrow \mathrm{I}=\frac{1}{2\sqrt{2}}\log \left|\frac{\sqrt{1+{\mathrm{t}}^2}-\sqrt{2}}{\sqrt{1+{\mathrm{t}}^2}+\sqrt{2}}\right|+\mathrm{C}=\frac{1}{2\sqrt{2}}\log \left|\frac{\sqrt{1+{\displaystyle \frac{1}{{\mathrm{x}}^2}}}-\sqrt{2}}{\sqrt{1+{\displaystyle \frac{1}{{\mathrm{x}}^2}}}+\sqrt{2}}\right|+\mathrm{C} \\ \Rightarrow \mathrm{I}=\frac{1}{2\sqrt{2}}\log \left|\frac{\sqrt{1+{\mathrm{x}}^2}-\sqrt{2}\mathrm{x}}{\sqrt{1+{\mathrm{x}}^2}+\sqrt{2}\mathrm{x}}\right|+\mathrm{C} \\ \left(3\right) \\ \mathrm{I}=\int \frac{{\mathrm{x}}^5}{\sqrt{1+{\mathrm{x}}^3}}\mathrm{dx}=\int \frac{{\mathrm{x}}^3.{\mathrm{x}}^2}{\sqrt{1+{\mathrm{x}}^3}}\mathrm{dx} \\ \mathrm{Let},1+{\mathrm{x}}^3=\mathrm{u} \\ \Rightarrow 3{\mathrm{x}}^2\mathrm{dx}=\mathrm{du}\Rightarrow {\mathrm{x}}^2\mathrm{dx}=\frac{1}{3}\mathrm{du} \\ \mathrm{Therefore}, \\ \mathrm{I}=\frac{1}{3}\int \frac{\mathrm{u}-1}{\sqrt{\mathrm{u}}}\mathrm{du}=\frac{1}{3}\int \left(\sqrt{\mathrm{u}}-\frac{1}{\sqrt{\mathrm{u}}}\right)\mathrm{du}=\frac{1}{3}\int \left({\mathrm{u}}^{\frac{1}{2}}-{\mathrm{u}}^{-\frac{1}{2}}\right)\mathrm{du} \\ \Rightarrow \mathrm{I}=\frac{1}{3}\left(\frac{{\mathrm{u}}^{{\displaystyle \frac{3}{2}}}}{{\displaystyle \frac{3}{2}}}-\frac{{\mathrm{u}}^{{\displaystyle \frac{1}{2}}}}{{\displaystyle \frac{1}{2}}}\right)+\mathrm{C}=\frac{1}{3}\left(\frac{2}{3}{\mathrm{u}}^{\frac{3}{2}}-2{\mathrm{u}}^{\frac{1}{2}}\right)+\mathrm{C}=\frac{2}{3}\sqrt{\mathrm{u}}\left(\frac{\mathrm{u}}{3}-1\right)+\mathrm{C} \\ \Rightarrow \mathrm{I}=\frac{2}{3}\sqrt{1+{\mathrm{x}}^3}\left(\frac{\sqrt{1+{\mathrm{x}}^3}}{3}-1\right)+\mathrm{C} \\ \left(4\right) \\ \mathrm{I}=\int \frac{\mathrm{x}}{\sqrt{1+{\mathrm{x}}^2}}\mathrm{dx} \\ \mathrm{Let},1+{\mathrm{x}}^2=\mathrm{v} \\ \Rightarrow 2\mathrm{x}\mathrm{dx}=\mathrm{dv} \\ \Rightarrow \mathrm{x}\mathrm{dx}=\frac{1}{2}\mathrm{dv} \\ \mathrm{Therefore}, \\ \mathrm{I}=\frac{1}{2}\int \frac{1}{{\mathrm{v}}^{{\displaystyle \frac{1}{2}}}}\mathrm{dv}=\frac{1}{2}\left(\frac{{\mathrm{v}}^{{\displaystyle \frac{1}{2}}}}{{\displaystyle \frac{1}{2}}}\right)+\mathrm{C}={\mathrm{v}}^{\frac{1}{2}}+\mathrm{C} \\ \Rightarrow \mathrm{I}=\sqrt{1+{\mathrm{x}}^2}+\mathrm{C} \\ \left(5\right) \\ \mathrm{I}=\int \frac{\mathrm{x}+1}{\sqrt{{\mathrm{x}}^2+1}}\mathrm{dx}=\int \frac{\mathrm{x}}{\sqrt{1+{\mathrm{x}}^2}}\mathrm{dx}+\int \frac{1}{\sqrt{1+{\mathrm{x}}^2}}\mathrm{dx} \\ \Rightarrow \mathrm{I}=\sqrt{1+{\mathrm{x}}^2}+\int \frac{1}{\sqrt{1+{\mathrm{x}}^2}}\mathrm{dx}\left[\mathrm{From}\left(4\right)\mathrm{above}\right] \\ \Rightarrow \mathrm{I}=\sqrt{1+{\mathrm{x}}^2}+\log \left|\mathrm{x}+\sqrt{1+{\mathrm{x}}^2}\right|+\mathrm{C}\left[\mathrm{Using}\int \frac{1}{\sqrt{{\mathrm{a}}^2+{\mathrm{x}}^2}}\mathrm{dx}=\log \left|\mathrm{x}+\sqrt{{\mathrm{a}}^2+{\mathrm{x}}^2}\right|+\mathrm{C}\right] \end{gathered}$$