Chapter IV


4.1 Direct integration

4.1.1 Basic rules of integration:

4.1.2 Table of standard Integrals:

Example 1. EXERCISES 1031 - 1050

Applying Basic Rules 1, 2, 3 and the formulae of integration, find the integrals:

Hints and Answers 1031 -1050

4.1.3 Integration under the sign of the differential: Rule 4) considerably expands the table of standard integrals: By virtue of this rule, the table of integrals holds true irrespective of whether the variable of integration is an independent variable or a differentiable function.

Example 2.

where we set u = 5x - 2. We have used Rule 4 and Integral I .

Example 3.

We set u=x and use Rule 4 and Integral V.

Example 4.

where we have used Rule 4 and Integral VII.

In Examples 2, 3, and 4, we have reduced the given integral to the following form before making use of a tabular integral:

This type of transformation is called integration under the differential sign. Common transformations of differentials, used .in Examples 2 and 3, are:

EXERCISES 1051 - 1144

Use the basic rules and integration formulae to find the integrals:

EXERCISES 1145 - 1190

Find the indefinite integrals:

Hints and Answers 1051 - 1190

4.2 Integration by Substitution

4.2.1 Change of variable in an indefinite Integral: Setting

where t is a new variable and j is a continuously differentiable function, we have

You aim to choose the function j in such a way that the right hand side of (1) becomes more convenient for integration.

Example 1. Find

Solution: It is natural to set t = (x - 1), whence x = t + 1 and dx = 2tdt. Thus,

At times, you use substitutions of the form

Suppose we have succeeded in transforming the integrand f(x)dx to the form

If you know


Actually, we have already uses this method in 1.3. Examples 2, 3, 4 of 4.1.2 may be solved as follows:

Example 2: u = 5x - 2, du = 2xdx, dx = du/2.

Example 3: u = x, du = 2xdx, xdx = du/2.

Example 4: u = x, du = 3x, xdx = du/3.

4.2.2 Trigonometric substitutions:

1) If an integral contains the root (a - x), it is standard to set x = a sin t, whence

2) If an integral contains the root (x - a), we set x = a sec t, whence 3)

3) If an integral contains the root (x+ a), we set x = tan t; whence

Note that trigonometric substitutions do not always turn out to be advantageous.

It is sometimes more convenient to make use of hyperbolic substitutions, which are similar to trigonometric substitutions (cf. Example 1209).

For more details about trigonometric and hyperbolic substitutions, see 4.9.

Example 5. Find

Solution: Set x = tan t, whence dx = dt/cost.

EXERCISES 1191 - 1210

Applying these substitutions, find the integrals:

1209. Find

by applying the hyperbolic substitution x = a sinh t.

Solution: We have




we finally get

where C1 = C - a/2 ln a is a new arbitrary constant.

Hints and Answers 1191 - 1210

4.3 Integration by Parts

Formula for Integration by parts. If u = j(x) and v = y(x) are differentiable functions, then

Example 1. Find

Setting u = ln x, dv = xdx, we have du = dx/x, v = x/2, whence

Sometimes, in order to reduce a given integral to a tabulated form, one has to apply the formula of integration by parts several times.. In certain cases, integration by parts will yield an equation by which the desired integral is determined.

Example 2. Find

We have



EXERCISES 1211 - 1254

Applying integration by parts, find the integrals:

Hints and Answers 1211 - 1254

4.4 Standard Integrals Containing a second order polynomial

4.4.1 Integrals of the form

The principal procedure is to reduce the quadratic to the form

where k and l are constants. In order to perform the transformations in (1), it is best to take the perfect square out of the polynomial. The following substitution may also be used:

If m = 0, then, reducing the quadratic to the form (1), we get the tabulated integrals III or IV .

Example 1.

If m 0, then we can take from the numerator the derivative 2ax + b out of the quadratic

and thus we arrive at the integral discussed above.

Example 2.

4.4.2 Integrals of the form

The methods are similar to those discussed above. The integral is finally reduced to the tabulated integral V, if a > 0, and VI, if a <0.

Example 3.

Example 4.

4.4.3 Integrals of the form
By the inverse substitution

these integrals are reduced to integrals of the form 4.4.2.

Example 5. Find

Solution: Set



4.4.4 Integrals of the form By taking the square out of the quadratic, the given integral is reduced to one of the two basic integrals (Examples 1252 and 1253):

Example 6.

EXERCISES 1259 - 1279

Find the integrals:

Hints and Answers 1255 - 1279

4.5 Integration of Rational Functions

4.5.1 Method of undetermined coefficients: Integration of a rational function reduces to integration of the rational fractions

where P(x) and Q(x) are polynomials and the degree of the numerator is lower than that of the denominator.


where a, . ...are real distinct roots of the polynomial Q(x) and a, ..., l are integers (root multiplicities), then the decomposition of (1) into partial fractions is possible:

In order to calculate the undetermined coefficients A1, A2, , both sides of Identity (2) are reduced to an integral form and then the coefficients of like powers of the variable x are equated (first method). These coefficients may likewise be determined by setting [in (2) or an equivalent equation] x equal to suitably chosen numbers (second method).

Example 1. Find

Solution: We have


a) First method: Rewrite (3) in the form

Equating the coefficients of identical powers of x, we get


b) Second method:. Setting x = 1 in (3), we find

Setting x = -l, we find

Finally, setting x = 0, we find



Example 2. Find

Solution: We have


When solving this example, it is advisable to combine the two methods of determining the coefficients. Applying the second method, we set x = 0 in identity (4) and find 1 = A. Then, setting x = l, we find 1= C. Moreover, by first method, we equate the coefficients of x in (4) and get

: whence,


If the polynomial Q (x) has complex roots a ib of multiplicity k, then partial fractions of the form

will enter into the (2). Here,

and A1, B2, ..., Ak, Bk are undetermined coefficients which must be determined by the methods given above. For k = 1, the fractions(5) is integrated directly; for k > 1, we use the reduction method, where it is at first advisable to represent the quadratic x2 + px + q in the form

and set x + p/2 = z.

Example 3. Find

Solution: Since

setting + 2 = z, we get

4.5.2 Ostrogradsky's method. If Q (x) has multiple roots, then

where Q1(x) is the largest common divisor of the polynomial Q(x) and its derivative

Q2(x) = Q(x) : Q1(x).

X(x) and Y(x) are polynomials with unknown coefficients, the degrees of which are, less by unity than those of Q1(x) and Q2(x), respectively.

The undetermined coefficients of the polynomials X(x) and Y(x) are computed by differentiating Identity (6).

Example 4. Find




Differentiating this identity, we get

Equating the coefficients of the respective degrees of x, we find


and, consequently,

In order to compute the integral on the right hand side of (7), we decompose the fraction into partial fractions:


Setting x = 1. we find L = 1/3.

Equating the coefficients of identical powers of x on the two sides (8), we find



EXERCISES 1280 - 1314

Find the integrals:

Hints and Answers 1280 - 1314

4.6 Integrating Certain Irrational Functions

4.6.1 Integrals of the form

where R is a rational function and p1, q1, p2, q2 are integers.

Such integrals are found by means of the substitution

where n is the least common multiple of the numbers q1, q2, .

Example 1. Find

Solution: The substitution 2x - 1 = z4' leads to an integral of the form

EXERCISES 1315 - 1325

Find the integrals

Hints and Answers 1315 - 1325

4.6.2. Integrals of the form
where Pn(x) is a polynomial of degree n.


where Qn-1(x) is a polynomial of degree (n - 1) with undetermined coefficients and h is a number. The coefficients of the polynomial Qn-1(x) and the number l. are found by differentiating (3).

Example 2.


Multiplying by (x + 4) and equating the coefficients of the same powers of x, we obtain


4.6.3 Integrals of the form are reduced to integrals of the form 4.6.2 by the substitution

EXERCISES 1326 - 1331

Hints and Answers 1326 - 1331

4.6.4 Integrals of the binomial differentials where m, n and p are rational numbers.

Chebyshev's conditions: The Integral (5) can be expressed in terms of a finite combination of elementary functions only in the three cases:

1) p is an integer,
2) (m + 1)/n is an integer, where we substitute a + bxn = zs with s the denominator of the fraction p,
3) (m + 1)/n + p is an integer, when we use substitute ax-n + b = xs.

Example 3. Find

Solution: Here,

whence, by 4.6.2, the substitution




EXERCISES 1332 - 1337


Hints and Answers 1332 - 1337

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