FUNDAMENTAL PRINCIPLES OF INTEGRATION - General Methods of Integration - The Calculus Primer

The Calculus Primer (2011)

Part XII. General Methods of Integration


12—3. Basic Relationships. From what has already been said, it will be seen that many integrations can be performed “by inspection.” If the reader has thoroughly mastered the fundamental formulas for derivatives, he will, with practice, be able to integrate many expressions at sight. However, we would wish to impress him with the need for having the fundamental formulas for differentiating, especially in differential notation, “at fingers’ ends.” Ready command of these formulas will go a long way toward success in integrating. Facility in integrating can be achieved only by diligent practice.

One of the basic principles for integrating is, of course, that

image dx = x + C, [1]

which follows at once from the definitions of the symbols and the meaning of the constant of integration.

Another principle is the following:

Since image

or d(av) = a dv;

therefore image a dv = a image dv. [2]

Or, expressing the role played by a constant multiplier in the integrand in another way, we have,

image av dx = a image v dx.

In words, a constant factor may be removed from one side of the integration sign to the other without affecting the value of the integral. This is a very useful procedure, but the reader should be warned that this can be done only with a constant, and never with a variable. For example:

image 2x3 dx = 2 image x3 dx.

image πa2x2 dx = πa2 image x2 dx.

A third basic principle is the following:

image (du + dv − dw) = image du + image dv − image dw, [3]

or, in different symbolism,

image {f(x) + ø(x) − F(x)} dx = image f(x) dx + image ø(x) dx − F(x) dx.

In words, the integral of the sum of any number of functions is equal to the sum of the integrals of the several functions. This follows, of course, from the fact that

EXAMPLE 1. Find image (ax2 + bx + c) dx.


image (ax2 + bx + c) dx = image ax2 dx + image bx dx + image c dx

= a image x2 dx + b image x dx + c image dx


EXAMPLE 2. Find image image


12—4. The Process of Integration. It is important to point out that the result of an integration does not always lead to a simple function. That is to say, performing an inverse operation in mathematics sometimes yields unusual results. Thus in arithmetic, when adding positive numbers only, the inverse operation of subtraction may lead to negative numbers, which differ in kind from the sort of numbers operated upon in arithmetic in the first place. Again, in algebra, when raising real numbers to the second power, the inverse operation of taking the square root may lead to imaginaries, a different kind of number. So in calculus: when we differentiate various so-called elementary functions (i.e., algebraic expressions resulting from the operations of addition, subtraction, multiplication, division, involution, evolution; exponential and logarithmic functions; and trigonometric functions and their inverses) the result of the differentiation contains only elementary functions; but the inverse operation of integration may yield a new kind of function, not an elementary function at all. However, in this book we shall limit ourselves to integrals which are elementary functions.

12—5. The Form image Un dU. The integral of a power of a variable, when the power is not −1, is given by the following:

This may be readily verified by reversing the procedure:

Thus, to integrate a variable to a given power, we increase the exponent by that amount and also divide the expression by the amount of the new exponent.

EXAMPLE 1. Find image x6 dx,


EXAMPLE 2. Find image ax dx.


EXAMPLE 3. Find image image


EXAMPLE 4. Find image x¾ dx.


EXAMPLE 5. Find image image


EXAMPLE 6. Find image (2a − 3x)2 dx.

Solution. Multiplying out first:

image (2a − 3x)2 dx = image (4a2 − 12ax + 9x2) dx

= 4a2 image dx − 12a image x dx + 9 image x2 dx


= 4a2x − 6ax2 + 3x3 + C.


Integrate each of the following; check your result by differentiating:

1. image 8x3 dx

2. image 10ax4 dx

3. image y4 dy

4. image x¾dx

5. image 2 image dx

6. image image x dx

7. image zimage dz

8. image 5y dy

9. image xm/n dx



15. image kz dz

16. image 2x−2 dx


23. image (x + 3)2 dx

24. image (3a − x2)2 dx

25. image 5(x + 2)3 dx

12—6. Variations of the Form image Un dU. At this point it might be well to suggest to the reader that there is no general or infallible method of finding the integral of a given function, since it is impossible always to retrace the steps of a differentiation. What we must do is this: depending upon how familiar we are with the derivative formulas, we try to recognize the given function as the derivative of some known function, or to change it to a form which can so be recognized. Every process of integration involves bringing the integrand to a form to which some standard, known formula applies. Considerable ingenuity is often required to do this.

EXAMPLE 1. Find image image.


Now, we may write dx = d(a + x).

Hence, the given integral may be written:

image (a + x)½ dx = image (a + x)½ d(a + x);

here, if we consider u = a + x, and n = image, then our integral can be found by means of equation [4]:

EXAMPLE 2. Find image image.

EXAMPLE 3. Find image (b2a2x2)½x dx.

Solution. We note that

d(b2 a2x2) = −2a2x dx.

Multiplying inside the integral sign by −2a2, and dividing outside by −2a2, leaves the integral unchanged in value, but transforms it to the form image un du, where u = (b2a2x2), and n = image;



EXAMPLE 4. Find image (x3 + 1)2x2 dx.

Solution. We can first multiply out:

image (x3 + 1)2x2 dx = image (x6 + 2x3 + 1)x2 dx

= image (x8 + 2x5 + x2)dx


Or, an alternative, more elegant solution. Since d(x3 + 1) = 3x2 dx, we may replace the “x2 dx” in the original problem by “imaged(x3 + 1)”; thus

We leave it to the reader to verify the fact that the right-hand members of (1) and (2), respectively, will both check when differentiated.


Integrate; check by differentiating:



14. image (m + x)(a + x) dx

15. Integrate image (x4 1)2x3 dx in two ways, explain why the results are not identical, and show that both are correct.

12—7. The Form image image. The integral of the reciprocal of a variable is the logarithm of the variable; or

This, of course, is readily verified by differentiation; thus,

NOTE. Strictly speaking, we should write image image = log |u| + C, since, although image is defined for all values of u ≠ 0, log u is defined only for values of u > 0.

EXAMPLE 1. Find image image.


EXAMPLE 2. Find image image

Solution. Since, from the definition of a differential, we may write, if we care to, dx = d(x + 1).

Hence image

where we are regarding u as equal to x + 1, and

image = 1, or du = 1(dx) = dx.

By differentiating the answer, the result is seen to check; thus

In fact, whenever the numerator of an integrand is (or can be made) the differential of the denominator, the integral is the logarithm of the denominator.

EXAMPLE 3. Find image image.

Solution. d(x2 + 1) = 2x.

Rewrite the given example:

now, considering u as x2 + 1, and image as 2x, or du = 2x dx, we have:

EXAMPLE 4. Find image image

Solution. First rewrite the integral as follows:


Find the following integrals; check by differentiation: