Question

⚠️Calcule a integral usando o método das frações parciais⚠️

Answer 1

Resposta:

∫ (x³+x+1)/(x²-1) dx

(x³+x+1)/(x²-1)

x³+x+1   |  x²-1

             |   x

-x³+x

+

2x+1

>>>>>>>>>>>> (x³+x+1)/(x²-1) = x  + (2x+1)/(x²-1)

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(2x+1)/(x²-1) = A/(x-1) + B/(x+1)

2x+1=A(x+1) +B(x-1)

2x+1=x*(A+B) +A-B

{A+B=2

{A-B=1

2A=3  ==>A=3/2

B=A-1=3/2-1=1/2

(2x+1)/(x²-1) = 3/2(x-1) + 1/2(x+1)

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(x³+x+1)/(x²-1) = x + 3/2(x-1) + 1/2(x+1)

∫ (x³+x+1)/(x²-1) dx

∫ x + 3/2(x-1) + 1/2(x+1) dx

∫ x dx=

x²/2  + c

∫ 3/2(x-1) dx

fazendo u=x-1 ==>du=dx

(3/2) *∫ 1/u du = (3/2) * ln|u| +c   ==>como u=x-1

=(3/2) * ln| x-1| + c

∫ 1/2(x+1) dx

fazendo u=x+1  ==>du=dx

(1/2) * ∫1/u  du = (1/2)* ln|u|  ==>como u=x+1

=(1/2) * ln(x+1| + c

Resposta é a soma das 3 parcelas

=  x²/2  +  (3/2) * ln| x-1| +(1/2) * ln(x+1| + c

Answer 2

$\int\limits\frac{x^3 + x + 1}{x^2-1}dx \\\\=\int \frac{x^3}{x^2-1}dx+\int \frac{x}{x^2-1}dx+\int \frac{1}{x^2-1}dx\\\\=\frac{1}{2}\left(x^2-1+\ln \left|x^2-1\right|\right)+\frac{1}{2}\ln \left|x^2-1\right|-\frac{1}{2}\ln \left|x+1\right|+\frac{1}{2}\ln \left|x-1\right|\\\\=\frac{1}{2}x^2-\frac{1}{2}+\ln \left|x^2-1\right|-\frac{1}{2}\ln \left|x+1\right|+\frac{1}{2}\ln \left|x-1\right|$$=\frac{1}{2}x^2-\frac{1}{2}+\ln \left|x^2-1\right|-\frac{1}{2}\ln \left|x+1\right|+\frac{1}{2}\ln \left|x-1\right +C\\\\=\frac{1}{2}x^2-\frac{1}{2}+\ln \left|x^2-1\right|-\ln \left(\left|x+1\right|^{\frac{1}{2}}\right)+\frac{1}{2}\ln \left|x-1\right|+C\\\\=\frac{1}{2}x^2-\frac{1}{2}+\ln \left(\frac{\left|x^2-1\right|}{\left|x+1\right|^{\frac{1}{2}}}\right)+\frac{1}{2}\ln \left|x-1\right|+C$$=\frac{1}{2}x^2-\frac{1}{2}+\ln \left(\frac{\left|x^2-1\right|\sqrt{\left|x+1\right|}}{\left|x+1\right|}\right)+\frac{1}{2}\ln \left|x-1\right|+C\\\\\\=\frac{1}{2}x^2-\frac{1}{2}+\ln \left(\frac{\left|x^2-1\right|\sqrt{\left|x+1\right|}}{\left|x+1\right|}\right)+\ln \left(\left|x-1\right|^{\frac{1}{2}}\right)+C\\\\=\frac{1}{2}x^2-\frac{1}{2}+\ln \left(\frac{\left|x^2-1\right|\sqrt{\left|x+1\right|}}{\left|x+1\right|}\left|x-1\right|^{\frac{1}{2}}\right)+C$$=\frac{1}{2}x^2-\frac{1}{2}+\ln \left(\frac{\left|x^2-1\right|\left|x-1\right|^{\frac{1}{2}}\sqrt{\left|x+1\right|}}{\left|x+1\right|}\right)+C\\\\=\frac{x^2}{2}-\frac{1}{2}+\ln \left(\frac{\left|x^2-1\right|\left|x-1\right|^{\frac{1}{2}}\sqrt{\left|x+1\right|}}{\left|x+1\right|}\right)+C\\\\=\frac{x^2-1}{2}+\ln \left(\frac{\left|x^2-1\right|\left|x-1\right|^{\frac{1}{2}}\sqrt{\left|x+1\right|}}{\left|x+1\right|}\right)+C$