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Integration

Integration by Substitution 2 - Harder Algebraic Substitution

We assume that you are familiar with the material in integration by substitution 1.

Recall the Substitution Rule

To integrate if we replaceby andby.

This converts the original integral into a simpler one. We extend this idea to some harder integrals.

Sometimes an integrand may need a bit of algebraic manipulation to make it integrable. Make sure that the integral with respect to the variable u does not contain any x.

Consider some special cases:

f(ax² + b)x^m dx

f(ax² + b)x^m dx where m is odd can be rewriiten as f(ax² + b)x^{m − 1} x dx

and the substitution u = ax² + b used.

Example

Let us consider x³(x² + 1)⁹⁹dx.

(x² + 1)⁹⁹ is a composite function. One would consider replacing the inner function (x² + 1) by u.

Thus du = 2x dx, which can be obtained from x³ since it equals x² times x.

We rewrite the integral as ½x²(x² + 1) ⁹⁹ 2x dx.

Since u = x² + 1, x² = u − 1

Hence the integral becomes ½(u − 1)u⁹⁹ du = ½(u¹⁰⁰ − u⁹⁹) du

This is integrable.

½(u¹⁰⁰ − u⁹⁹) du = ½((1/101)u¹⁰¹ −(1/100)u¹⁰⁰) + c = (1/202)(x² + 1)¹⁰¹ − (1/200)(x² + 1)¹⁰⁰ + c

Exercise

Evaluate the indefinite integral: dx =

Integrand contains (mx + b)^(m/n)

When the integrand contains an expression of the form (mx + b)^(m/n) then the substitution u = (mx + b)^(1/n) is often suitable. Thus uⁿ = mx + b and nu^{n − 1} du = m dx.

Example

Evaluate .

The square root is a composite function with inner function 2x + 1. Let u = which is equivalent to u² = 2x + 1

Then 2u du = 2 dx or u du = dx. We need to express the factor x in terms of u.

u² = 2x + 1 so x = ½(u² − 1)

We can write the integral as

Hence the integral becomes

Integrating and substituting for u:

Exercises

dx =

<< Integration by Substitution 1 | Integration Index | Integration by Trigonometric Substitution 1 >>