Integral indefinida

por gsr_principiamathematica Qui 28 Dez 2023, 15:42

Resolva a integral [latex]\int arcsin(\sqrt{\frac{x}{x+1}})dx[/latex].

por **Giovana Martins** Qui 28 Dez 2023, 18:47

Acredito que seja isto.

[latex]\mathrm{Integrando\ por\ partes:\int udv=uv-\int vdu}[/latex]

[latex]\mathrm{u=arcsin\left ( \frac{\sqrt{x}}{\sqrt{x+1}} \right ),dv=1\ \therefore\ du=\frac{\frac{1}{2\sqrt{x}\sqrt{x+1}}-\frac{\sqrt{x}}{2(x+1)^{\frac{3}{2}}}}{\sqrt{1-\frac{x}{x+1}}},v=x}[/latex]

[latex]\mathrm{\int arcsin\left ( \sqrt{\frac{x}{x+1}} \right )dx=xarcsin\left ( \frac{\sqrt{x}}{\sqrt{x+1}} \right )-\int \frac{\frac{x}{2\sqrt{x}\sqrt{x+1}}-\frac{x\sqrt{x}}{2(x+1)^{\frac{3}{2}}}}{\sqrt{1-\frac{x}{x+1}}}dx[/latex]

[latex]\mathrm{\int \frac{\frac{x}{2\sqrt{x}\sqrt{x+1}}-\frac{x\sqrt{x}}{2(x+1)^{\frac{3}{2}}}}{\sqrt{1-\frac{x}{x+1}}}dx=\frac{1}{2}\int \frac{\sqrt{x}}{x+1}dx}[/latex]

[latex]\mathrm{t=\sqrt{x}\ \therefore\ dx=2\sqrt{x}dt\ \therefore\ \int \frac{\sqrt{x}}{x+1}dx=2\int \frac{t^2}{t^2+1}dt=2\int \left ( \frac{t^2+1}{t^2+1}-\frac{1}{t^2+1} \right )dt}[/latex]

[latex]\mathrm{\int \frac{\sqrt{x}}{x+1}dx=\int (1)dt-\int \frac{1}{t^2+1}dt=t-arctan(t)=2\sqrt{x}-2arctan\left ( \sqrt{x} \right )}[/latex]

[latex]\mathrm{\boxed{\therefore\mathrm{\int arcsin\left ( \sqrt{\frac{x}{x+1}} \right )dx=xarcsin\left ( \sqrt{\frac{x}{x+1}} \right )-\sqrt{x}+arctan\left ( \sqrt{x} \right )+C}}}[/latex]

Integral indefinida

Integral indefinida

Re: Integral indefinida

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