Question

Integral de raiz (1 + x^2) multiplicado por x^5 dx

Answer 1

Calcular a integral indefinida:

     $\displaystyle\int\sqrt{1+x^2}\cdot x^5\,dx\\\\\\ =\int \sqrt{1+x^2}\cdot x^4\cdot x\,dx\\\\\\ =\int(1+x^2)^{1/2}\cdot (x^2)^2\cdot x\,dx$


Embora seja possível resolvê-la por substituição trigonométrica, podemos fazer outra substituição:

     $(1+x^2)^{1/2}=u\\\\\\ \Rightarrow\quad\left\{\!\begin{array}{lcl}1+x^2=u^2&\quad \Rightarrow\quad&x^2=u^2-1\\\\ 2x\,dx=2u\,du&\quad\Rightarrow\quad&x\,dx=u\,du \end{array}\right.$


Substituindo, a integral fica

     $\displaystyle=\int u\cdot (u^2-1)^2\cdot u\,du\\\\\\ =\int u^2\cdot (u^2-1)^2\,du$


Expanda o quadrado da diferença:

     $\displaystyle=\int u^2\cdot (u^4-2u^2+1)\,du\\\\\\ =\int (u^6-2u^4+u^2)\,du$

     $=\dfrac{u^{6+1}}{6+1}-2\cdot \dfrac{u^{4+1}}{4+1}+\dfrac{u^{2+1}}{2+1}+C\\\\\\ =\dfrac{u^7}{7}-2\cdot \dfrac{u^5}{5}+\dfrac{u^3}{3}+C\\\\\\ =\dfrac{1}{7}\,u^7-\dfrac{2}{5}\,u^5+\dfrac{1}{3}\,u^3+C$


Substitua de volta para a variável $x:$

     $=\dfrac{1}{7}\,[(1+x^2)^{1/2}]^7-\dfrac{2}{5}\,[(1+x^2)^{1/2}]^5+\dfrac{1}{3}\,[(1+x^2)^{1/2}]^3+C$

     $=\dfrac{1}{7}\,(1+x^2)^{7/2}-\dfrac{2}{5}\,(1+x^2)^{5/2}+\dfrac{1}{3}\,(1+x^2)^{3/2}+C\quad\longleftarrow\quad\textsf{resposta.}$


Bons estudos! :-)