integration by partial fractions

Let us learn about integration by partial fractions

If the integrand is in the form of an algebraic fraction & the integral cannot be evaluated by simple methods, the fraction requires to be expressed in partial fractions before integration takes place.

We decompose fractions into partial fractions like this due it makes certain integrals much easier to do, & it is applied in the Laplace transform, which we meet later.

Evaluate ∫ (x² + 1) / (x² -5x + 6) dx.

The integration by partial fractions is not a proper rational function on dividing

(x² + 1) by (x² - 5x + 6), we get

(x² + 1) / (x² - 5x + 6) =1 + (5x - 5) / (x² -5x + 6) =1 + (5x -5) / (x - 2) (x - 3)

Now, let (5x - 5) / ( x - 2)(x - 3) = A / ( x -2) + B / ( x -3)

=> (5x - 5) / (x - 2) (x -3) = A(x -3) + B(x -2) / (x -2) (x -3)

Placing x =2 on both sides of (i), we get A = -5

Placing x =3 on both sides of (i), we get B =10

( x² + 1) / (x² - 5x +6) =1 - 5 / (x -2) + 10 / (x -3)

=> ∫ (x² + 1) / (x² -5x + 6)dx = ∫ dx - 5 ∫ dx / (x - 2) + 10 ∫ dx / (x -3)

= x -5 log | x -2 | + 10 log | x -3 | + C

In our next blog we shall learn about extraction of aluminium I hope the above explanation was useful.Keep reading and leave your comments.