INTEGRATE QUADRATIC FUNCTION IN THE DENOMINATOR

We have to express ax2+bx+c as sum or difference of two square terms to get the integrand in one of the standard forms.

Example 1 :

Solution :

We should represent the quadratic equation which is in the denominator in the form of sum or difference of squares.

Using completing the square method, we get

=  x2 + 2 ⋅x ⋅ (5/2)  + (5/2)- (5/2)2+7

=  (x+(5/2))2+7-25/4

=  (x+(5/2))2+3/4

=  (x+(5/2))2+(3/2)2

This exactly matches the formula,

∫1/(x2+a2) dx  =  1/a tan-1(x/a) + c

Here a  =  √3/2 and x  = x+(5/2)

Example 2 :

Solution :

Using completing the square method, we get

=  x2 - 2 ⋅x ⋅ (7/2)  + (7/2)- (7/2)2+5

=  x2 - 2 ⋅x ⋅ (7/2)  + (7/2)- (49/4)+5

=  (x-(7/2))2+5-49/4

=  (x-(7/2))2-29/4

=  (x-(7/2))2-(√29/2)2

This exactly matches the formula,

∫1/(x2-a2) dx  =  (1/2a) [log (x-a)/(x+a)] + c

Here a  =  √29/2 and x  = x-(7/2)

Integrating Quadratic Function in the Denominator with Square Root

Example 3 :

Solution :

Using completing the square method, we get 

=  x2 + 2⋅x⋅8 + 82-82+100

=  (x+8)2 -64+100

=  (x+8)2+36

=  (x+8)2+62

This exactly matches the formula,

∫1/(a2+x2) dx  =  log[x+(a2+x2)] + c

Here a  =  6 and x  = x+8

Example 4 :

Solution :

Using completing the square method, we get 

9+8x-x =  -(x2-8x-9)

=  -(x2 - 2⋅x⋅4 + 42-42-9)

=  -[(x-4)2-16-9]

=  -[(x-4)2-25]

=  52-(x-4)2

This exactly matches the formula,

∫1/(a2-a2) dx  =  sin-1(x/a) + C

Here a  =  5 and x  =  x-4

Integration of Rational Functions by Partial Fractions

Example 5 :

Using partial fraction, we are decomposing

∫(4x-3)/(x2+3x+8) dx 

Numerator  =  A (Derivative of denominator) + B  --(1)

4x-3  =  A d(x2+3x+8)/dx + B

4x-3  =  A (2x+3) + B

Equating the coefficients of x, we get

4  =  2A

A  =  2

Equating constants, we get

-3  =  3A + B

By applying the value of A, we get

-3  =  6 + B

B  =  -9

Numerator  =  A (Derivative of denominator) + B

Dividing (1) by (x2+3x+8), we get

(4x-3)/(x2+3x+8) dx

=  ∫2 (2x+3)/(x2+3x+8) dx - 9/(x2+3x+8) dx

=  2∫[(2x+3)/(x2+3x+8)]dx - 9∫[1/(x2+3x+8)] dx

By integrating 2∫[(2x+3)/(x2+3x+8)]dx :

By integrating 9∫1/(x2+3x+8)dx :

Using completing the square method, we get

=  x2 + 2⋅x⋅(3/2) + (3/2)2-(3/2)2+8

=  (x+(3/2))2 + 8 - 9/4

=  (x+(3/2))2 + 23/4

=  (x+(3/2))2 √(23/2)2

The second part exactly matches with the formula

∫1/(x2+a2) dx  =  1/a tan-1(x/a) + c

Here x  =  x+(3/2) and a  =  √(23/2)

Using the above formula, we get

=  1/√(23/2) tan-1 (2x+3)/23

=  (2/√23) tan-1 (2x+3)/√23

By combing the above results, we get

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