Question

ache $\int\dfrac{x^2}{\sqrt{x^2+x+1}}dx$​

Answer 1

Explicação passo a passo:

Calcular a integral indefinida

    $\displaystyle\int \frac{x^2}{\sqrt{x^2+x+1}}\,dx$

Devemos reescrever a expressão do segundo grau de dentro da raiz quadrada como uma soma ou diferença de quadrados. Para isso, completamos os quadrados aplicando os produtos notáveis. O termo constante 1 foi reescrito como a soma 1/4 + 3/4:

    $\displaystyle=\int \frac{x^2}{\sqrt{(x^2+x+\frac{1}{4})+\frac{3}{4}}}\,dx\\\\\\\\=\int \frac{x^2}{\sqrt{(x+\frac{1}{2})^2+(\frac{\sqrt{3}}{2})^2}}\,dx$

Chamemos u = x + 1/2, e a constante a = √3/2. Então,

    $x=u-\dfrac{1}{2}\quad\Longrightarrow\quad dx=du$

Substituindo, a integral fica

    $\displaystyle =\int\frac{(u-\frac{1}{2})^2}{\sqrt{u^2+a^2}}\,du\qquad (i)$

Agora, temos que fazer uma substituição trigonométrica. Façamos

   $u=a\,\mathrm{tg\,}t\quad\Longrightarrow\quad du=a\sec^2 t\,dt,\qquad ~\mathrm{com~}-\dfrac{\pi}{2}</p><p>Com isso, temos</p><p></p><p>     [tex]\Longrightarrow\quad u^2+a^2=(a\,\mathrm{tg\,}t)^2+a^2\\\\\Longrightarrow\quad u^2+a^2=a^2\,\mathrm{tg^2\,}t+a^2\\\\\Longrightarrow\quad u^2+a^2=a^2(\mathrm{tg^2\,}t+1)\\\\\Longrightarrow\quad u^2+a^2=a^2\sec^2 t\\\\\Longrightarrow\quad \sqrt{u^2+a^2}=a\sec t$

Substituindo em (i), a integral fica

    $\displaystyle =\int\frac{(a\,\mathrm{tg\,}t-\frac{1}{2})^2}{a\sec t}\cdot a\sec^2 t\,dt\\\\\\=\int \Big(a\,\mathrm{tg\,}t-\frac{1}{2}\Big)^2\cdot \sec t\,dt\\\\\\=\int \Big(a^2\,\mathrm{tg^2\,}t-a\,\mathrm{tg\,}t+\frac{1}{4}\Big)\cdot \sec t\,dt\\\\\\=a^2\int \mathrm{tg^2\,}t\sec t\,dt-a\int \mathrm{tg\,}t\sec t\,dt+\frac{1}{4}\int \sec t\,dt\qquad (ii)$

Vamos avaliar a primeira integral da expressão acima:

    $\displaystyle\int \mathrm{tg^2\,}t\sec t\,dt\\\\\\=\int (\sec^2 t-1)\sec t\,dt$

Integrando por partes:

    $\begin{array}{lcl} v=\sec t&\quad\Longrightarrow\quad&dv=\mathrm{tg\,}t\sec t\,dt\\\\dw=(\sec^2 t-1)\,dt&\quad\Longleftarrow\quad&w=\mathrm{tg\,}t-t\end{array}$

Logo,

    $\displaystyle\int \mathrm{tg^2\,}t\sec t\,dt=\sec t(\mathrm{tg\,}t-t)-\int (\mathrm{tg\,}t-t)\,\mathrm{tg\,}t\sec t\,dt\\\\\\\Longrightarrow\quad \int \mathrm{tg^2\,}t\sec t\,dt=\sec t\,\mathrm{tg\,}t-t\sec t-\int\mathrm{tg^2\,}t\sec t\,dt+\int t\,\mathrm{tg\,}t\sec t\,dt\\\\\\\Longrightarrow\quad 2\int \mathrm{tg^2\,}t\sec t\,dt=\sec t\,\mathrm{tg\,}t-t\sec t+\int t\,\mathrm{tg\,}t\sec t\,dt$

    $\displaystyle \Longrightarrow\quad \int \mathrm{tg^2\,}t\sec t\,dt=\frac{1}{2}\sec t\,\mathrm{tg\,}t-\frac{1}{2}\,t\sec t+\frac{1}{2}\int t\,\mathrm{tg\,}t\sec t\,dt\qquad (iii)$

Integrando por partes novamente:

    $\begin{array}{lcl} v=t&\quad\Longrightarrow\quad&dv=dt\\\\dw=\mathrm{tg\,}t\sec t\,dt&\quad\Longleftarrow\quad&w=\sec t\end{array}$

Substituindo em (iii):

    $\displaystyle \Longrightarrow\quad \int \mathrm{tg^2\,}t\sec t\,dt=\frac{1}{2}\sec t\,\mathrm{tg\,}t-\frac{1}{2}\,t\sec t+\frac{1}{2}\left(t\sec t-\int\sec t\,dt\right)\\\\\\\Longrightarrow\quad \int \mathrm{tg^2\,}t\sec t\,dt=\frac{1}{2}\sec t\,\mathrm{tg\,}t-\frac{1}{2}\,t\sec t+\frac{1}{2}\,t\sec t-\frac{1}{2}\int\sec t\,dt\\\\\\\Longrightarrow\quad \int \mathrm{tg^2\,}t\sec t\,dt=\frac{1}{2}\sec t\,\mathrm{tg\,}t-\frac{1}{2}\ln |\sec t+\mathrm{tg\,}t|+C_1$

Voltando a (ii), a integral fica

    $\displaystyle =a^2\left(\frac{1}{2}\sec t\,\mathrm{tg\,t}-\frac{1}{2}\ln |\mathrm{tg\,}t+\sec t|\right)-a\int \mathrm{tg\,}t\sec t\,dt+\frac{1}{4}\int \sec t\,dt\\\\\\ =a^2\left(\frac{1}{2}\sec t\,\mathrm{tg\,t}-\frac{1}{2}\ln |\mathrm{tg\,}t+\sec t|\right)-a\sec t+\frac{1}{4}\ln |\sec t+\mathrm{tg\,}t|+C_2\\\\\\ =\frac{a^2}{2}\sec t\,\mathrm{tg\,}t-a\sec t+\left(-\,\frac{a^2}{2}+\frac{1}{4}\right)\ln |\sec t+\mathrm{tg\,}t|+C_2\qquad (iv)$

Agora, temos que voltar à variável x. Substituindo de volta sec t e tg t, e o valor da constante a, obtemos

    $=\dfrac{a^2}{2}\cdot \dfrac{\sqrt{u^2+a^2}}{a}\cdot \dfrac{u}{a}-a\cdot \dfrac{\sqrt{u^2+a^2}}{a}+\left(-\,\dfrac{a^2}{2}+\dfrac{1}{4}\right)\ln\left|\dfrac{\sqrt{u^2+a^2}}{a}+\dfrac{u}{a}\right|+C_2$

    $=\dfrac{1}{2}\,u\sqrt{u^2+a^2}-\sqrt{u^2+a^2}+\left(-\,\dfrac{a^2}{2}+\dfrac{1}{4}\right)\!\ln\left|\dfrac{\sqrt{u^2+a^2}+u}{a}\right|+C_2$  

   $=\left(\dfrac{1}{2}\,u-1\right)\!\sqrt{u^2+a^2}+\left(\dfrac{1-2a^2}{4}\right)\!\ln\left|\sqrt{u^2+a^2}+u\right|-\left(\dfrac{1-2a^2}{4}\right)\!\ln a+C_2$

    $=\left(\dfrac{1}{2}\,u-1\right)\!\sqrt{u^2+a^2}+\left(\dfrac{1-2a^2}{4}\right)\!\ln\left|\sqrt{u^2+a^2}+u\right|+C$

    $=\left[\dfrac{1}{2}\Big(x+\dfrac{1}{2}\Big)-1\right]\!\sqrt{x^2+x+1}-\dfrac{1}{8}\ln\left|\sqrt{x^2+x+1}+\Big(x+\dfrac{1}{2}\Big)\right|+C\\\\\\ =\left(\dfrac{1}{2}\,x-\dfrac{3}{4}\right)\!\sqrt{x^2+x+1}-\dfrac{1}{8}\ln\left|\sqrt{x^2+x+1}+\Big(x+\dfrac{1}{2}\Big)\right|+C$

    $=\dfrac{1}{2}\Big(x-\dfrac{3}{2}\Big)\sqrt{x^2+x+1}-\dfrac{1}{8}\ln\left|\sqrt{x^2+x+1}+\Big(x+\dfrac{1}{2}\Big)\right|+C$

Dúvidas? Comente.

Bons estudos!