Question

Calcule a integral.$\int\limits^, {\frac{x}{\sqrt{x^2+x+1} } } \, dx$

Answer 1

Olá, siga a explicação:

"Dicas relevantes"

Sendo a integral indefinida:

$\sf \displaystyle \int\limits \dfrac{x}{\sqrt{x^2 + x + 1} }$

Completa o quadrado:

$\boxed { \sf x^2+x+1 : & \left ( x+ \dfrac{1}{2} \right ) ^2 + \dfrac{3}{4} }$

$\sf = \displaystyle \int\limits \dfrac{x}{\sqrt{ \left ( x+ \dfrac{1}{2} \right ) ^2+ \dfrac{3}{4} } } dx$

Integral por Substituição:

$\boxed {\sf u= x + \dfrac{1}{2} }$

$\sf = \displaystyle \int\limits \dfrac{2u -1}{\sqrt{4u^2 + 3} } du$

Expande:

$\boxed { \sf \dfrac{2u - 1}{\sqrt{4u^2 + 3} } : & \dfrac{2u}{\sqrt{4u^2 + 3} } - \dfrac{1}{\sqrt{4u^2 + 3} } }$

$\sf = \displaystyle \int\limits \dfrac{2u}{\sqrt{4u^2 + 3} } - \dfrac{1}{\sqrt{4u^2 + 3} } du$

Regras De Soma:

$\boxed {\sf \displaystyle \int\limits f(x) \pm g(x) dx= \displaystyle \int\limits f(x) dx \pm \displaystyle \int\limits g(x) dx}$

$\sf = \displaystyle \int\limits \dfrac{2u}{\sqrt{4u^2 + 3} } du - \displaystyle \int\limits \dfrac{1}{\sqrt{4u^2 + 3} } du$

$\boxed { \sf \displaystyle \int\limits \dfrac{2u}{\sqrt{4u^2 + 3} } du = \dfrac{1}{2} \sqrt{4u^2 + 3} }$

$\boxed { \sf \displaystyle \int\limits \dfrac{1}{\sqrt{4u^2 + 3} } du = \dfrac{1}{2} In \left | \dfrac{2}{\sqrt{3} } u + \sqrt{\dfrac{1}{3} \left ( 3+ 4u^2 \right ) } \right | }$

$\sf = \dfrac{1}{2} \sqrt{4u^2 + 3} - \dfrac{1}{2} In \left | \dfrac{2}{\sqrt{3}}u+\sqrt{\dfrac{1}{3}\left(3+4u^2\right)} \right |$

Substituir na equação:

$\boxed {\sf u= x +\dfrac{1}{2} }$

$\sf \dfrac{1}{2}\sqrt{4\left(x+\dfrac{1}{2}\right)^2+3}-\dfrac{1}{2}\ln \left|\dfrac{2}{\sqrt{3}}\left(x+\dfrac{1}{2}\right)+\sqrt{\dfrac{1}{3}\left(3+4\left(x+\dfrac{1}{2}\right)^2\right)}\right|$

Simplifica:

$\boxed { \sf \dfrac{1}{2}\sqrt{4\left(x+\dfrac{1}{2}\right)^2+3}-\dfrac{1}{2}\ln \left|\dfrac{2}{\sqrt{3}}\left(x+\dfrac{1}{2}\right)+\sqrt{\dfrac{1}{3}\left(3+4\left(x+\dfrac{1}{2}\right)^2\right)}\right| }$

$= \sf \sqrt{x^2+x+1}-\dfrac{1}{2}\ln \left|\dfrac{2}{\sqrt{3}}x+\dfrac{1}{\sqrt{3}}+\sqrt{\dfrac{4x^2+4x+4}{3}}\right|$

$\boxed { = \sf \sqrt{x^2+x+1}-\dfrac{1}{2}\ln \left|\dfrac{2}{\sqrt{3}}x+\dfrac{1}{\sqrt{3}}+\sqrt{\dfrac{4x^2+4x+4}{3}}\right|+C }$

• Att. MatiasHP

• Gráfico, Assumindo C=0:
• Tópicos De Resolver Uma Integral: