`int (x^2+6x+4)/(x^4+8x^2+16)dx`
To solve using partial fraction method, the denominator of the integrand should be factored.
`(x^2+6x+4)/(x^4+8x^2+16) = (x^2+6x+4) / (x^2+4)^2`
If the factor in the denominator is quadratic and repeating, the partial fraction of this factor is ` (A_1x+B_1)/(ax^2+bx+c)+(A_2x+B_2)/(ax^2+bx+c)^2 + ... +(A_nx+B_n)/(ax^2+bx+C)^n` .
So expressing the integrand as sum...
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of fractions, it becomes:
`(x^2+6x+4) / (x^2+4)^2=(Ax+B)/(x^2+4) + (Cx+D)/(x^2+4)^2`
To solve for the values of A, B, C and D, multiply both sides by the LCD.
`(x^2+4)^2 * (x^2+6x+4) / (x^2+4)^2=((Ax+B)/(x^2+4) + (Cx+D)/(x^2+4)^2) * (x^2+4)^2`
`x^2+6x+4=(Ax + B)(x^2+4) + Cx + D`
`x^2+6x + 4 = Ax^3 + 4Ax + Bx^2 + 4B + Cx + D`
At the right side, group together the terms with same power of x.
`x^2+6x+4 =Ax^3 + Bx^2 + (4Ax + Cx) + (4B + D)`
`x^2+6x+4=Ax^3 + Bx^2 + (4A+C)x + (4B + D)`
Notice that the right side has a degree of 3. So express the polynomial at the left side with a degree of 3.
`0x^3+x^2+6x+4=Ax^3 + Bx^2 + (4A+C)x + (4B + D)`
In order that the two polynomials to be equal, the coefficients and the constant should be the same.
So set the coefficient of `x^3` at the left side equal to the coefficient of `x^3` at the right side.
`0=A`
Also, set the coefficient of `x^2`at the left side equal to the coefficient of `x^2` at the right side.
`1=B`
Set the coefficient of x at the left side equal to the coefficient of x at the right side too.
`6=4A + C` (Let this be EQ1.)
And set the constant at the left side equal to the constant at the right side.
`4=4B+D` (Let this be EQ2.)
Since the values of A and B are known already, plug-in them to equation 1 and 2 to get the values of C and D.
Plug-in A=0 to EQ1 to get the value of C.
`6=4(0) +C`
`6=C`
And, plug-in B = 1 to EQ2 to get the value of D.
`4=4(1)+D`
`4=4+D`
`0=D`
So the partial fraction decomposition of the integrand is:
`int (x^2+6x+4)/(x^4+8x^2+16)dx`
= `int(x^2+6x+4) / (x^2+4)^2dx`
`=int (1/(x^2+4) + (6x)/(x^2+4)^2)dx`
Expressing it as sum of two integrals, it becomes:
`= int 1/(x^2+4)dx + int (6x)/(x^2+4)^2 dx`
`= int 1/(x^2+4)dx + 6int (x)/(x^2+4)^2 dx`
For the first integral, apply the formula `int 1/(u^2+a^2) du = 1/a tan^(-1) (u/a) + C` .
`u = x`
`du = dx`
`a=2`
For the second integral, apply the formula `int u^n du = u^(n+1)/(n+1)+C` .
`u = x^2+4`
`du = 2x dx`
So the result of each integral is:
`= int 1/(x^2+4)dx + 6int (x^2+4)^(-2) *xdx`
`= int 1/(x^2+4)dx + 3int (x^2+4)^(-2) *2xdx`
`= 1/2 tan^(-1)(x/2) + 3*(x^2+4)^(-1)/(-1)+C`
`= 1/2 tan^(-1)(x/2) - 3(x^2+4)^(-1)+C`
`= 1/2 tan^(-1)(x/2) - 3/(x^2+4)+C`
Therefore, `int (x^2+6x+4)/(x^4+8x^2+16)dx= 1/2 tan^(-1)(x/2) - 3/(x^2+4)+C` .