Question

(20 PONTOS) Calcule a primitiva:
$\displaystyle\int{\mathrm{arctg}{(\sqrt{x})\,dx}}$
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Resposta: $(x+1)\,\mathrm{arctg}(\sqrt{x})-\sqrt{x}+C.$

Answer 1

primeiro vc usa a integração por substituição e depois a integração por partes. segue em anexo como eu resolvi.

Answer 2

$\large\boxed{\begin{array}{l}\displaystyle\rm\int arctg(\sqrt{x})\,dx\\\rm z=\sqrt{x}\longrightarrow x=z^2\\\rm dx=2zdz\\\displaystyle\rm\int arctg(\sqrt{x})\,dx=\int arctg(z)\,2zdz\\\underline{\sf fac_{\!\!,}a}\\\rm u=arctg(z)\longrightarrow du=\dfrac{1}{z^2+1}dz\\\rm dv= 2z\,dz\longrightarrow v=z^2\\\displaystyle\rm\int arctg(z)2zdz= z^2arctg(z)-\int\dfrac{z^2}{z^2+1}dz\\\rm ~~z^2+0z+0|\underline{ z^2+0z+1}\\\rm\underline{-z^2~+0-1}~~1~~\\\rm ~~~~-1\\\rm\dfrac{z^2}{z^2+1}=1-\dfrac{1}{z^2+1}\end{array}}$

$\large\boxed{\begin{array}{l}\displaystyle\rm\int\dfrac{z^2}{z^2+1}dz=\int dz-\int\dfrac{dz}{z^2+1}=z-arctg(z)+k\\\\\displaystyle\rm\int arctg(\sqrt{z})\,2zdz= z^2arctg(z)-[z-arctg(z)]+k\\\\\displaystyle\rm\int arctg(z)\,2zdz= z^2arctg(z)-z+artctg(z)+k\\\sf retornando\,a\,substituic_{\!\!,}\tilde ao~original\,temos\\\displaystyle\rm\int arctg(\sqrt{x})\,dx=x\cdot arctg(\sqrt{x})-\sqrt{x}+arctg(\sqrt{x})+k\end{array}}$