- #1

SclayP

- 27

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So I'm stuck resolving this integral...

[itex]\int^{1}_{0} \sqrt{e^{2x} + 1} \, dx[/itex]

[tex]u=2x ; du=2dx[/tex]

[itex]\frac{1}{2}\int^{2}_{0} \sqrt{e^u + 1} \, du[/itex]

[tex]t=e^u+1 ; dt=e^u du[/tex]

[itex]\frac{1}{2}\int^{e^2 + 1}_{2} \frac{\sqrt{t}}{t-1} \, dt[/itex]

[tex]v=\sqrt{t} ; dv=\frac{1}{2\sqrt{t}}[/tex]

[itex]\int \frac{v^3}{v^2 -1} \, dv[/itex]

And here i don't know how to proceed, i red i have to do long division but i don't know why... please help me.

Thanks.

[itex]\int^{1}_{0} \sqrt{e^{2x} + 1} \, dx[/itex]

[tex]u=2x ; du=2dx[/tex]

[itex]\frac{1}{2}\int^{2}_{0} \sqrt{e^u + 1} \, du[/itex]

[tex]t=e^u+1 ; dt=e^u du[/tex]

[itex]\frac{1}{2}\int^{e^2 + 1}_{2} \frac{\sqrt{t}}{t-1} \, dt[/itex]

[tex]v=\sqrt{t} ; dv=\frac{1}{2\sqrt{t}}[/tex]

[itex]\int \frac{v^3}{v^2 -1} \, dv[/itex]

And here i don't know how to proceed, i red i have to do long division but i don't know why... please help me.

Thanks.

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