Integration of Irrational Functions | Mathematics (Maths) for JEE Main & Advanced PDF Download

Definition

Certain types of integrals of algebraic irrational expressions can be reduced to integrals of rational functions by a appropriate change of the variable. Such transformation of an integral is called its rationalization.

1. If the integrand is a rational function of fractional powers of an independent variable x, i.e. the function Rthen the integral can be rationalized by the substitution x = t^m, where m is the least common multiple of the numbers q₁, q₂, ...., q_k.
2. If the integrand is a rational function of x and fractional powers of a linear fractional function of the form  then rationalization of the integral is effected by the substitution  where m has the same sense as above.

Solved Examples

Ex.1 Evaluate  

Sol.

Rationalizing the denominator, we have  

Ex.2 Evaluate I = 

Sol. The least common multiple of the numbers 3 and 6 is 6, therefore we make the substitution

x = t⁶, dx = 6t⁵ dt.

Ex.3 Evaluate I = 

Sol. The integrand is a rational function of  therefore we put 2x – 3 = t⁶, whence

Returning to x, we get

Ex.4 Evaluate 

Sol.

Let x = t³ ⇒ dx = 3t² then

Ex.5 Evaluate I = 

Sol. The integrand is a rational function of x and the expression   therefore let us introduce the substitution

Integrals of the type   where x & y are linear or quadratic expressions

Ex.1 Integrate 

Sol.

Put 4x + 3 = t², so that 4dx = 2tdt and (2x + 1)  

Ex.2 Evaluate  

Sol.

Put (x + 2) = t², so that dx = 2t dt, Also x = t² – 2.

∴  

  dividing the numerator by the denominator

Ex.3 Integrate  

Sol.

Put (x + 1) = t², so that dx = 2t dt. Also x = t² – 1.

Ex.4 Integrate 

Sol.

Put (1 + x) = 1/t, so that dx = – (1/t²) dx.

Also x = (1/t) – 1.

Ex.5 Evaluate  

Sol.

Put x = 1/t, so that dx = – (1/t²) dt.

∴     

Now put 1 + t² = z² so that t dt = z dz. Then

 [∵ t = 1/x]

Ex.6 Evaluate I = 

Sol. 

 Integration Of A Binomial Differential

The integral  where m, n, p are rational numbers, is expressed through elementary functions only in the following three cases :

Case I : p is an integer. Then, if p > 0, the integrand is expanded by the formula of the binomial; but if p < 0, then we put x = t^k, where k is the common denominator of the fractions and n.

Case II : is an integer. We put a + bxⁿ = t^α, where α is the denominator of the fraction p.

Case III :+ p is an integer we put a + bxⁿ = t^αxⁿ, where a is the denominator of the fraction p.

Ex.1 Evaluate I = 

Sol.

  Here p = 2, i.e. an integer, hence we have case I.

Ex.2 Evaluate I = 

Sol. 

 i.e. an integer.we have case II. Let us make the substitution. Hence ,

Ex.3 Evaluate I = 

Sol. 

Here p = – 1/2 is a fraction, m+1/2 = -5/2 also a fraction, but m+1/n + p/2 = -5/2 -1/2 = -3 is an integer, i.e. we have case III, we put 1 + x⁴ = x^4/2,

Hence  

Substituting these expression into the integral, we obtain

Returning to x, we get I =