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:

**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* = 5*x* - 2. We
have used Rule 4 and Integral
I .

**Example 3. **

We set *u=x*² 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:

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

Find the indefinite integrals:

**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* = Ö(*x* - 1), whence *x = t*² + 1 and *dx
= *2*tdt.* 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

then

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
= *5*x - *2*, du *= 2*xdx, dx = du*/2.

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

**Example 4: ***u*
= *x*³, *du = *3*x*², *x*²*dx =
du*/3.

**4.2.2
Trigonometric substitutions****:**

1) If an integral contains the root Ö(*a*²*
- x*²)*,* it is standard to set *x* = *a *sin
*t*, whence

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

3) If an integral contains the root Ö(*x*²*+ a*²),
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*/cos²*t.*

Applying these substitutions, find the integrals:

1209. Find

by applying the hyperbolic
substitution *x* = *a* sinh *t*.

Solution: We have** **

whence

Since

and

we finally get

where *C*_{1} = C - *a*²/2
ln *a* is a new arbitrary constant.**
**

*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*²/2*,
*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

Hence,

whence

Applying integration by parts, find the integrals:

**4.4 Standard Integrals Containing a second order
polynomial**

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

and thus we arrive at the integral discussed above.

**Example 2.**

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

whence

Thus,

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.**

Find the integrals:

**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.

If

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 *A*_{1}*,**A*_{2},
···, 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**

whence

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

Equating the coefficients of identical powers
of *x,* we get

whence

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

Setting *x *= -l, we find

Finally, setting *x* = 0,
we find

i.e.,

Hence,

**Example 2.** Find

**Solution:** We have

and

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,

Thus,

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 *A*_{1}**,** *B*_{2},
..., *A*_{k}*,* *B*_{k}
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 *x*^{2} + *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 Q_{1}(x) is the largest common
divisor of the polynomial Q(*x*) and its derivative

**Q**_{2}(**x**)
= **Q**(**x**) : **Q**_{1}(**x**).

*X*(*x**)* and *Y*(*x)*
are polynomials with unknown coefficients, the degrees of which
are, less by unity than those of *Q*_{1}(*x*)
and Q_{2}(*x*), respectively.

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

**Example** **4.** Find

**Solution:**

Differentiating this identity, we get

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

whence

and, consequently,

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

i.e.,

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

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

Thus,

and

Find the integrals:

**4.6 Integrating Certain Irrational Functions**

where R is a rational function and *p*_{1}*,
q*_{1}*, p*_{2}*, q*_{2}*
*are integers.

Such integrals are found by means of the substitution

where *n* is the least common multiple of
the numbers *q*_{1}, *q*_{2},
···**.**

**Example**** 1.** Find

**Solution: **The substitution 2*x*
- 1 = *z*^{4}' leads to an integral of the form

Find the integrals

**4.6.2.
Integrals of the form ****
**where

Set

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

**Example 2. **

Thus,

Multiplying by Ö(*x*² + 4) and
equating the coefficients of the same powers of *x*, we
obtain

whence

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

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

**Chebyshev's
condition****s:**
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 + bx*^{n}* = z*^{s}*
*with s the denominator of the fraction *p*,

3) (*m + *1)/*n* + *p *is an integer, when
we use substitute *ax*_{-n} *+ b = x*^{s}*.*

**Example** **3.** Find

**Solution: **Here,

whence, by 4.6.2, the substitution

yields

and

where