Integration of Quadratic Polynomials    ax2  + bx  + c

 

 

In a Nut Shell:  For more complicated cases involving quadratic polynomials it

 may be necessary to split the integral into several integrals.  The example below

 illustrates this situation.

 

 

 

Example:    I   =   ∫ (2x + 3)dx / (9x2  + 6x  + 5)

 

Start by taking the derivative of the quadratic polynomial,  9x2  + 6x  + 5

 

Dx [9x2  + 6x  + 5 ] = 18 x+ 6

 

Then set      2x + 3  =  A(18x + 6) + B   so A = 1/9  and  B = 7/3

 

Therefore  2x + 3  =  1/9 (18x + 6)  +  7/3

 

Now split the original integral into two simpler integrals    I  =  I1  +  I2

where 

 

I1  =  ∫ 1/9[ (18x + 6) /(9x2  + 6x  + 5)]dx ,       I2 = 7/3 ∫ dx /(9x2  + 6x  + 5) 

 

This first integral is of the form  ∫du/u which results in the logarithmic expression

 

                              I1  =  1/9  ln |9x2  + 6x  + 5| + C1

 

 

 

For the second integral complete the square which gives

 

      9x2  + 6x  + 5 =  (3x + 1)2  +  22  and  I2 = 7/3 ∫ dx / [ (3x + 1)2  +  22 ]

 

  Then let u  =  3x + 1  so that  du  =  3 dx   or   dx  =  (1/3) du;  The integral becomes

 

                I2 = 7/3 ∫ (1/3) du / [ u2  +  22 ]  and use the trig substitution   u  =  2 tan θ

 

    So        du  =  2 sec2 θ  d θ  and    u2 + 22  =   22 sec2 θ

 

 and the second integral becomes   I2 = 7/9 ∫ ½  d θ   =  7/18  θ  + C2

                              

But   θ  =  tan-1 (u/2)  =  tan-1[ (3x+1)/2]  and therefore

 

I  =   1/9  ln |9x2  + 6x  + 5| + (7/18) + tan-1[ (3x+1)/2]  +  C    (result)

 

 

 


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