`int (x^2 + x + 1)/(x^2 + 1)^2 dx` Evaluate the integral

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`int (x^2+x+1)/(x^2+1)^2 dx`

To solve, apply partial fraction decomposition.

To express the integrand as sum of proper rational functions, set the equation as follows:


Multiply both sides by the LCD.

`x^2+x+1=(Ax+B)(x^2+1) + Cx + D`



Express the left side as a polynomial with degree 3.


For the two sides to be equal, the two polynomials should be the same. So set the coefficients of the polynomials equal to each other.


`0=A`     (Let this be EQ1.)


`1=B `     (Let this be EQ2.)


`1=A+C`     (Let this be EQ3.)


`1=B+D`     (Let this be EQ4.)

In the equations, the values of A and B are already known. So only the values of C and D have to be solved. To do so, plug-in the value of A to EQ3. 




Also, plug-in the value of B to EQ4.




So the partial fraction decomposition of the integrand is:


Taking the integral of it result to:

`int (x^2+x+1)/(x^2+1)^2 dx`

`= int (1/(x^2+1)+x/(x^2+1)^2) dx`

`= int 1/(x^2+1)dx + int x/(x^2+1)^2dx`

For the first integral, apply the formula int 1/(u^+a^2)=1/a tan^(-1) u/a+C.

For the second integral, apply u-substitution method.


     `du=2x dx`


`= int 1/(x^2+1)dx + int 1/u^2 * (du)/2`

`= int 1/(x^2+1)dx +1/2 int u^(-2) du`

`= tan^(-1)x -1/2u^(-1)+C`

`= tan ^(-1)x - 1/(2u)+C`

Substitute back `u=x^2+1` .

`= tan ^(-1)x - 1/(2(x^2+1))+C`


Therefore, `int (x^2+x+1)/(x^2+1)^2dx= tan ^(-1)x - 1/(2(x^2+1))+C` .

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