`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:
`(x^2+x+1)/(x^2+1)^2=(Ax+B)/(x^2+1)+(Cx+D)/(x^2+1)^2`
Multiply both sides by the LCD.
`x^2+x+1=(Ax+B)(x^2+1) + Cx + D`
`x^2+x+1=Ax^3+Bx^2+Ax+B+Cx+D`
`x^2+x+1=Ax^3+Bx^2+(A+C)x+B+D`
Express the left side as a polynomial with degree 3.
`0x^3+x^2+x+1=Ax^3+Bx^2+(A+C)x+B+D`
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.
x^3:
`0=A` (Let this be EQ1.)
x^2:
`1=B ` (Let this be EQ2.)
x:
`1=A+C` (Let this be EQ3.)
Constant:
`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.
`1=A+C`
`1=0+C`
`1=C`
Also, plug-in the value of B to EQ4.
`1=B+D`
`1=1+D`
`0=D`
So the partial fraction decomposition of the integrand is:
`(x^2+x+1)/(x^2+1)^2=(0x+1)/(x^2+1)+(1x+0)/(x^2+1)^2=1/(x^2+1)+x/(x^2+1)^2`
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.
`u=x^2+1`
`du=2x dx`
`(du)/2=xdx`
`= 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` .
Comments
Post a Comment