How do you find the indefinite integral of #int (x^2+2x+3)/(x^2+3x^2+9x)#?

Question

Christopher Agresta · Accepted Answer

#int(x^2+2x+3)/(x^2+3x^2+9x)dx=1/4x+1/16ln(4x+9)-2/3ln(|1/sqrt((8/9x+1)^2-1)+1/(8/9x+1)|)#

#I=int(x^2+2x+3)/(x^2+3x^2+9x)dx# #=int(x^2+2x+3)/(4x^2+9x)dx=1/4int(4x^2+8x+12)/(4x^2+9x)dx# #=1/4int((4x^2+9x)+x+12)/(4x^2+9x)dx#

#=1/4int1dx+1/4int(x+12)/(4x^2+9x)dx#

#=1/4x+1/4intx/(4x^2+9x)dx+1/4int12/(4x^2+9x)dx#

#=1/4x+1/16int4/(4x+9)dx+3/4int1/(x^2+9/4x)dx#

For the first integral, let #X=4x+9# #dX=4dx#

So:

#I=1/4x+1/16int1/XdX+3/4int1/((x+9/8)^2-81/64)dx#

Now let #x+9/8=9/8sec(theta)#

#dx=9/8sec(theta)tan(theta)d theta#, and because #sec(theta)^2-1=tan(theta)^2#::

#I=1/4x+1/16ln(X)+3/4int(9/8sec(theta)tan(theta))/(81/64tan(theta)^2)d theta#

#=1/4x+1/16ln(X)+2/3intsec(theta)/tan(theta)d theta#

#=1/4x+1/16ln(X)+2/3int(1/cos(theta))/(sin(theta)/(cos(theta)))d theta#

#=1/4x+1/16ln(X)+2/3int1/sin(theta)d theta#

We have #int1/sin(theta)d theta=-ln(|cot(theta)+csc(theta)|)#, you can have the proof here.

#I=1/4x+1/16ln(X)-2/3ln(|cot(theta)+csc(theta)|)#

#=1/4x+1/16ln(4x+9)-2/3ln(|cot(theta)+csc(theta)|)#

Because #theta=sec^(-1)(8/9x+1)#, #cot(theta)=cos(theta)csc(theta)#, #cos(sec^(-1)(x))=1/x# and #csc(sec^(-1)(x))=1/sin(sec^(-1)(x))=x/sqrt(x^2-1)# (here is a proof that #sin(sec^(-1)(x))=sqrt(x^2-1)/x#.),

#I=1/4x+1/16ln(4x+9)-2/3ln(|1/sqrt((8/9x+1)^2-1)+1/(8/9x+1)|)#

\0/ This is our response!

Emma Aldridge · Answer

undefined To find the indefinite integral of $\int \frac{x^2+2x+3}{x^2+3x^2+9x} \, dx$, you first need to factor the denominator.

$$x^2+3x^2+9x = x^2(1+3x+9)$$
$$= x^2(3x^2+3x+3)$$
$$= x^2(3)(x^2+x+1)$$

So the integral becomes:

$$\int \frac{x^2+2x+3}{x^2(3)(x^2+x+1)} \, dx$$

Now, we can use partial fraction decomposition. We have:

$$\frac{x^2+2x+3}{x^2(3)(x^2+x+1)} = \frac{A}{x} + \frac{B}{x^2} + \frac{Cx+D}{x^2+x+1}$$

Solving for $A$, $B$, $C$, and $D$, then integrating each term separately will yield the indefinite integral.