Question

Avalie a integral $\displaystyle\mathsf{\int arctg(\sqrt{x})~dx}$

Answer 1

Integral por Partes

$\large \boxed{\boxed{\boxed{\boxed{\displaystyle\mathsf{\int u\cdot dv=u\cdot v-\int v\cdot du}}}}}$

$\dotfill$

$\displaystyle\mathsf{\int arctg(\sqrt{x})~dx}$

faça

$\mathsf{t=\sqrt{x}\implies~x=t^2}\\\mathsf{dx=2t~dt}$

$\displaystyle\mathsf{\int arctg(\sqrt{x})~dx=\int arctg(t) 2t~dt}$

faça

$\mathsf{u=arctg(t)}\\\mathsf{du=\dfrac{dt}{1+t^2}}\\\mathsf{dv=2t~dt\implies v=t^2}$

$\displaystyle\mathsf{\int arctg(t) 2t~dt=arctg(t)\cdot t^2-\int \dfrac{t^2}{1+t^2}~dt}$

$\mathsf{\dfrac{t^2}{1+t^2}=1-\dfrac{1}{1+t^2}}$

$\displaystyle\mathsf{\int\dfrac{t^2}{1+t^2}~dt=\int dt-\int\dfrac{1}{1+t^2}~dt}\\\mathsf{=t-arctg(t)+c}$

$\displaystyle\mathsf{\int arctg(t) 2t~dt=arctg(t)\cdot t^2-[t-arctg(t)]+k}\\\displaystyle\mathsf{\int arctg(t)~dt=t^2arctg(t)-t+arctg(t)+k}$

Daí

$\displaystyle\mathsf{\int arctg(\sqrt{x})~dx} \\\mathsf{=x\cdot arctg(\sqrt{x})-\sqrt{x}+arctg(\sqrt{x})+k}$

Answer 2

Oie, Td Bom?!

$\displaystyle\int \: arctan( \sqrt{x} ) \: dx$

$\displaystyle\int \: arctan( \sqrt{x} ) \: . \: \frac{1}{ \frac{1}{2 \sqrt{x} } } \: dt$

$\displaystyle\int \: arctan( \sqrt{x} ) \: . \: 2 \sqrt{x} \: dt$

$\displaystyle\int \: 2arctan( \sqrt{x} ) \: . \: \sqrt{x} \: dt$

• Substitua $\sqrt{x} = t$.

$\displaystyle\int \: arctan(t) \: . \: t \: dt$

$2 \: . \: \displaystyle\int \: arctan(t) \: . \: t \: dt$

• Integrando por partes definindo:

• (I) u.

$u = arctan(t)$

$du = \frac{1}{1 + t {}^{2} } \: dt$

• (II) dv.

$dv = t \: dt$

$v = \frac{t {}^{2} }{2}$

• Substituindo $u = arctan(t)$,

$v = \frac{t {}^{2} }{2}$, $du = \frac{1}{1 + t {}^{2} } \: dt$ e $dv = t \: dt$ em $\displaystyle\int \: u \: dv = uv - \displaystyle\int \: v \: du$ .

• Continuando, então teremos:

$2(arctan(t) \: . \: \frac{t {}^{2} }{2} - \displaystyle\int \: \frac{t {}^{2} }{2} \: . \: \frac{1}{1 + t {}^{2} } \: dt$

$2(arctan(t) \: . \: \frac{t {}^{2} }{2} - \displaystyle\int \: \frac{t {}^{2} }{2(1 + t {}^{2}) } \: dt$

$2(arctan(t) \: . \: \frac{t {}^{2} }{2} - \frac{1}{2} \: . \: \displaystyle\int \: \frac{t {}^{2} }{1 + t {}^{2} } \: dt)$

$2(arctan(t) \: . \: \frac{t {}^{2} }{2} - \frac{1}{2} \: . \: \displaystyle\int \: \frac{t {}^{2} + 1 - 1}{1 + t {}^{2} } \: dt)$

$2(arctan(t) \: . \: \frac{t {}^{2} }{2} - \frac{1}{2} \: . \: \displaystyle\int \: \frac{t {}^{2} + 1}{ 1 + t {}^{2} } - \frac{1}{1 + t {}^{2} } \: dt)$

$2(arctan(t) \: . \: \frac{t {}^{2} }{2} - \frac{1}{2} \: . \: (\displaystyle\int \frac{t {}^{2} + 1}{1 + t {}^{2} } \: dt - \int \frac{1}{1 + t {}^{2} } \: dt))$

$2(arctan(t) \: . \: \frac{t {}^{2} }{2} - \frac{1}{2} \: . \: (t - arctan(t)))$

• Substitua $t = \sqrt{x}$.

$2(arctan( \sqrt{x} ) \: . \: \frac{ \sqrt{x} {}^{2} }{2} - \frac{1}{2} \: . \: ( \sqrt{x} - arctan( \sqrt{x} )))$

$2(arctan( \sqrt{x} ) \: . \: \frac{x}{2} - \frac{1}{2} \: . \: \sqrt{x} + \frac{arctan( \sqrt{x} )}{2} )$

$2( \frac{arctan( \sqrt{x} ) \: . \: x}{2} - \frac{1}{2} \: . \: \sqrt{x} + \frac{arctan( \sqrt{x}) }{2} )$

$2( \frac{arctan( \sqrt{x} ) \: . \: x + arctan( \sqrt{x}) }{2} - \frac{1}{2} \: . \: \sqrt{x} )$

$arctan( \sqrt{x} ) \: . \: x + arctan( \sqrt{x} ) - \sqrt{x}$

$arctan( \sqrt{x} ) \: . \: x + arctan( \sqrt{x} ) - \sqrt{x} + C$

Att. Makaveli1996