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**Integration of Irrational Functions**

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.

(i) 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 q_{1}, q_{2}, ...., q_{k}.

(ii) 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.

**Ex.68 Evaluate **

**Sol.**

Rationalizing the denominator, we have

**Ex.69 Evaluate I = **

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

x = t^{6}, dx = 6t^{5} dt.

**Ex.70 Evaluate I = **

**Sol.** The integrand is a rational function of therefore we put 2x â€“ 3 = t^{6}, whence

Returning to x, we get

**Ex.71 Evaluate **

**Sol.**

Let x = t^{3} â‡’ dx = 3t^{2} then

**Ex.72 Evaluate I = **

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

**INTEGRAL OF THE TYPE ** **WHERE X AND Y ARE LINEAR OR QUADRATIC EXPRESSION**

**Ex.73 Integrate **

**Sol.**

Put 4x + 3 = t^{2}, so that 4dx = 2tdt and (2x + 1)

**Ex.74 Evaluate **

**Sol.**

Put (x + 2) = t^{2}, so that dx = 2t dt, Also x = t^{2 }â€“ 2.

âˆ´

dividing the numerator by the denominator

**Ex.75 Integrate **

**Sol.**

Put (x + 1) = t^{2}, so that dx = 2t dt. Also x = t^{2} â€“ 1.

**Ex.76 Integrate **

**Sol.**

Put (1 + x) = 1/t, so that dx = â€“ (1/t^{2}) dx.

Also x = (1/t) â€“ 1.

**Ex.77 Evaluate **

**Sol.**

Put x = 1/t, so that dx = â€“ (1/t^{2}) dt.

âˆ´

Now put 1 + t^{2} = z^{2} so that t dt = z dz. Then

[âˆµ t = 1/x]

**Ex.78 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 + bx^{n} = t^{Î±}, where Î± is the denominator of the fraction p.

**Case III :**+ p is an integer we put a + bx^{n} = t^{Î±}x^{n}, where a is the denominator of the fraction p.

**Ex.79 Evaluate I = **

**Sol.**

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

**Ex.80 Evaluate I = **

**Sol. **

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

**Ex.81 Evaluate I = **

**Sol. **

Here p = â€“ 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 + x^{4} = x^{4/2},

Hence

Substituting these expression into the integral, we obtain

Returning to x, we get I =

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