Question

Integral x+1/(x-1)^4 dx

Answer 1

Integral de funções racionais por frações parciais com fatores lineares repetidos.

$\frac{x + 1}{ {(x - 1)}^{4} } = \frac{A}{x - 1} + \frac{B}{ {(x - 1)}^{2}}+\frac{C}{ {(x - 1)}^{3}} + \frac{D}{ {(x - 1)}^{4} }$

$\frac{x + 1 = A {(x - 1)}^{3}+B {(x - 1)}^{2} +C {(x - 1)}^{3}+D {(x - 1)}^{4} }{ {(x - 1) }^{4}}$

$x + 1 = A {(x - 1)}^{3}+B {(x - 1)}^{2} +C(x - 1) \\ + D$

$x + 1 = A {x}^{3} - 3A {x}^{2}+3Ax - A \\+B {x}^{2}-2Bx+B +Cx -C +D$

$x + 1 = A {x}^{3} + (-3A+B){x}^{2} \\+(3A - 2B + C)x \\ - A + B - C + D$

Pela identidade de polinômios temos

$\begin{cases}\mathsf{A=0}\\\mathsf{-3A+B=0}\\\mathsf{3A-2B+C=1}\\\mathsf{-A+B-C+D=1}\end{cases}$

$\boxed{\boxed{\mathsf{A=0}}}$

$\boxed{\boxed{\mathsf{-3.0+B=0\to~B=0}}}$

$\boxed{\boxed{\mathsf{C=1}}}$

$\mathsf{-1+D=1}\\\mathsf{D=1+1}$

$\boxed{\boxed{\mathsf{D=2}}}$

$\displaystyle\mathsf{\int\dfrac{x+1}{{(x-1)}^{4}}} \\ = \displaystyle\mathsf{\int\dfrac{x}{{(x-1)}^{4}}dx} + \displaystyle\mathsf{\int\dfrac{2}{{(x-1)}^{4}}dx}$

$\mathsf{u=x-1\to~x=u+1}\\\mathsf{du=dx} \\ \displaystyle\mathsf{\int\dfrac{x}{{(x-1)}^{4}}dx} = \displaystyle\mathsf{\int\dfrac{u + 1}{{u}^{4}}du}$

$\displaystyle\mathsf{\int({u}^{-3}+{u}^{-4})du = - \frac{1}{2{u}^{2} } - \frac{1}{ 3{u}^{3 }} + k}$

$\displaystyle\mathsf{\int\dfrac{x}{{(x-1)}^{4}}dx} \\ = \mathsf{ -\frac{1}{2 {(x - 1)}^{2}} - \frac{1}{3( {x - 1})^{3} } + k}$

$\mathsf{t=x-1\to~dt=dx}\\\displaystyle\mathsf{\int\dfrac{2}{{(x-1)}^{4}}dx= \displaystyle \mathsf{\int\dfrac{2}{{t}^{4}}dt}} \\\mathsf{ - \frac{2}{ 3{t}^{3} } + k }$

$\mathsf{t=x-1\to~dt=dx}\\\displaystyle\mathsf{\int\dfrac{2}{{(x-1)}^{4}}dx=\displaystyle \mathsf{\int\dfrac{2}{{t}^{4}}dt}} \\=\mathsf{-\frac{2}{3 {t}^{3}} + k}$

$\displaystyle\mathsf{\int\dfrac{2}{{(x-1)}^{4}}dx}=\mathsf{\dfrac{-2}{3{(x-1)}^{3}}+k}$

$\displaystyle\mathsf{\int\dfrac{x+1}{{(x-1)}^{4}}}=\\-\dfrac{1}{2{(x-1)}^{2}}-\dfrac{1}{3{(x-1)}^{3}}-\dfrac{2}{3{(x-1)}^{3}}</p><p>.$

$\displaystyle\mathsf{\int\dfrac{x+1}{{(x-1)}^{4}}} = \\ \mathsf{ -\frac{1}{2 {(x - 1)}^{2} } - \frac{1}{ {(x - 1)}^{3} } + k }$