Student Question

# int (x^2+6x+4)/(x^4+8x^2 + 16) dx Use partial fractions to find the indefinite integral

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^2at 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 .

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