Question

Nu reușesc să rezolv următoarea integrală. Ceva idei? ​Cerința e să se afle primitiva funcției f.

Answer 1

$\int _{\:}^{\:}\:\frac{x^4+1}{x^3+1}dx=\int \:\frac{x^4+x-x+1}{x^3+1}dx=\int \:\frac{x\left(x^3+1\right)}{x^3+1}dx-\int \:\frac{x-1}{x^3+1}dx=\int \:xdx-\int \frac{x-1}{x^3+1}dx$

$\frac{x-1}{x^3+1}=\frac{x-1}{\left(x+1\right)\left(x^2-x+1\right)}=\frac{A}{x+1}+\frac{Bx+C}{x^2-x+1}$

$x-1=A\left(x^2-x+1\right)+\left(Bx+C\right)\left(x+1\right)$$x-1=Ax^2-Ax+A+Bx^2+Cx+Bx+C$

$x-1=x^2\left(A+B\right)+x\left(B+C-A\right)+A+C$

$A=-\frac{2}{3},\:B=\frac{2}{3},\:C=-\frac{1}{3}$

$\frac{x-1}{x^3+1}=-\frac{2}{3\left(x+1\right)}+\frac{2x-1}{3\left(x^2-x+1\right)}$

$\int \:\frac{x-1}{x^3+1}dx=\int \:\frac{2x-1}{3\left(x^2-x+1\right)}dx-\int \:\frac{2}{3\left(x+1\right)}dx=\frac{1}{3}\int \left(\ln \left(x^2-x+1\right)\right)^'}dx-\frac{2}{3}\int \:\left(\ln \left(x+1\right)\right)^'}dx=$

$=\frac{1}{3}\ln \left(x^2-x+1\right)-\frac{2}{3}\ln \left(x+1\right)+C$

$\int \:\frac{x^4+1}{x^3+1}dx=\int \:xdx-\int \:\:\frac{x-1}{x^3+1}dx=\frac{x^2}{2}-\frac{1}{3}\ln \:\left(x^2-x+1\right)+\frac{2}{3}\ln \:\left(x+1\right)+C$