Methods of Integration

The standard formulae for integration are only useful when the integrand is given in the ‘standard’ form. For most physical applications or analysis purposes, advanced techniques of integration are required, which reduce the integrand analytically to a suitable solvable form. Two such methods – Integration by Parts, and Reduction to Partial Fractions are discussed here. Both have limited applicability, but are of immense use in solving integrands which may seem unsolvable otherwise. So let’s begin!

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Integration by Parts

Another useful technique for evaluating certain integrals is integration by parts. Here, the integrand is usually a product of two simple functions (whose integration formula is known beforehand). One of the functions is called the ‘first function’ and the other, the ‘second function’. The general formula for the Integration by parts method then can be given as – $$ \int uv dx = u\int v dx – \int[\frac{d}{dx}(u)\int vdx]dx + c $$
where both u and v are functions of x.
u(x) – the first function
v(x) – the second function

The preference for deciding the first and second functions is usually as follows –
Remember the code ILATE for the precedence, where
I – Inverse functions
L – Logarithmic Functions
A – Algebraic Functions
T – Trigonometric Functions
E – Exponential Functions

Application – Suppose your integrand is a product of two functions – exponential and logarithmic. On comparing with the ILATE form of precedence, the logarithmic function will be chosen as the first function and the exponential function can then be taken as the second function for easy evaluation. Thus, for solving ∫(e^x)log(4x²)dx, we can choose log(4x²) as the first function, and the e^x as the second function to easily obtain the result.

Browse more Topics under Calculus

• Application of Marginal Cost And Marginal Revenue
• Introduction to Integral Calculus

Method of Partial Fractions

This method relies on the fact that the integration of functions of the form \(\frac{1}{f(x)}\), where f(x) is a linear function with some exponent, can be done quite easily. Thus, the integrands involving polynomial functions in their numerator and denominator are reduced to partial fractions first, to ease the process of integration.

Different types of integrands need to be handled differently under this method. Therefore, we have the following three general cases –

Assume that the integrand is of the form \(\frac{P(x)}{Q(x)}\) with both P(x) and Q(x) being polynomials that may be factorized into multiple polynomials. Under every case, we’ll evaluate the partial fractions only under the condition that the degree of P(x) is less than the degree of Q(x).

Case I: Q(x) contains non-repeated linear factors

Basically, Q(x) in this case should be of the form (x – a)(x – b)(x – c)… while P(x) may be any polynomial that satisfies the condition of its degree being less than that of Q(x).

Example – To find the partial fractions corresponding to \(\frac{x}{(2 – x)(x + 3)}\):
Let $$\frac{x}{(2 – x)(x + 3)} = \frac{A}{(2 – x)} + \frac{B}{(x + 3)}$$ $$= \frac{A(x + 3) + B(2 – x)}{(2 – x)(x + 3)}$$

To find A and B, now we will equate the coefficients of the numerators on both sides of the equation, since the fraction represented by them is the same. We then get –

x = A(x + 3) + B(2 – x)

Now we can proceed by equating the coefficients of equal powers of x on both sides, which is the standard method of solving such algebraic equations; or, we may use a shortcut.

Note that on substituting x = -3 in the equation, we get –

-3 = A(-3 + 3) + B(2 – (-3))
-3 = 0 + 5B
B = -3/5

Similarly, on substituting x = 2;

2 = A(2 + 3) + B(2 – 2)
2 = 5A + 0
A = 2/5

Clearly, all we did was to subsitute the values of x that made the coefficient of A and B = 0 separately. Our final result then can be written as -$$\frac{x}{(2 – x)(x + 3)} = \frac{A}{(2 – x)} + \frac{B}{(x + 3)}$$ $$\frac{x}{(2 – x)(x + 3)} = \frac{2}{5(2 – x)} – \frac{5}{3(x + 3)}$$

You can also see from the shortcut that you may directly substitute the values of x for which the denominators of our partial fractions tend to go to 0 as well. For eg – To make the denominator of \(\frac{B}{(x + 3)}\) = 0, we substitute x = -3 throughout the equation, after multiplying the entire equation by the denominator of  \(\frac{B}{(x + 3)}\) i.e. (x + 3). We’ll then be left with –

\(\frac{x}{(2 – x)(x + 3)}.(x + 3) = \frac{A}{(2 – x)}.(x + 3) + \frac{B}{(x + 3)}.(x + 3)\)
\(\frac{x}{(2 – x)} = \frac{A}{(2 – x)}.(x + 3) + B\)
Now put x = -3 to get –
\(\frac{3}{(2 – (-3))} = \frac{A}{(2 – x)}.(-3 + 3) + B\)
\(B = -\frac{5}{3}\), which is the same result as before.

A similar analysis can be done to find A as well. This way, all the partial fractions of an integrand of the form \(\frac{P(x)}{Q(x)}\) can be found, where the degree of P(x) is less than that of Q(x).

Case II: Q(x) contains one or more repeated linear factors

Here, Q(x) may have repeated roots and could be of the form (x – a)(x – b)³(x – c)²…
In such a case, if a linear factor is repeated $n$ times in the denominator, there will be $n$ corresponding partial fractions with degree 1 to $n$.

Example – To find the partial fractions corresponding to \(\frac{1}{(x – 1)^2(x + 1)}\):
Let $$\frac{1}{(x – 1)^2(x + 1)} = \frac{A}{(x – 1)} + \frac{B}{(x – 1)^2} + \frac{C}{(x + 1)}$$ $$= \frac{A(x – 1)(x + 1) + B(x + 1) + C(x – 1)^2}{(x – 1)^2(x + 1)}$$

To find A,B and C, now we will equate the coefficients of the numerators on both sides of the equation, since the fraction represented by them is the same. We then get –

1 = A(x – 1)(x + 1) + B(x + 1) + C(x – 1)²

Now we can proceed by equating the coefficients of equal powers of x on both sides or, once again we may proceed by direct substitution.

Note that on substituting x = 1 in the equation, we get –

1 = A(1 – 1)(1 + 1) + B(1 + 1) + C(1 – 1)²
1 = 0 + 2B + 0
B = 1/2

Similarly, on substituting x = -1;

1 = A(-1 – 1)(-1 + 1) + B(-1 + 1) + C(-1 -1)²
1 = 0 + 0 + 4C
C = 1/4

Now to get A, we may substitute the values of B and C in the numerator balance equation, along with any arbitrary value of x (besides 1 and -1) to get –

1 = A(x – 1)(x + 1) + B(x + 1) + C(x – 1)²
Put B = 1/2 and C = 1/4
1 = A(x – 1)(x + 1) + \(\frac{1}{2}\)(x + 1) + \(\frac{1}{4}\)(x – 1)²
Then let us put x = 0 in this case because it seems like an easy value to evaluate the equation for.
: 1 = A(0 – 1)(0 + 1)  + \(\frac{1}{2}\)(0 + 1) + \(\frac{1}{4}\)(0 – 1)² 
: 1 = -A + \(\frac{1}{2}\) + \(\frac{1}{4}\)
: A = -\(\frac{1}{4}\)

Our final result then can be written as -$$\frac{1}{(x – 1)^2(x + 1)} = \frac{A}{(x – 1)} + \frac{B}{(x – 1)^2} + \frac{C}{(x + 1)}$$ $$ = – \frac{1}{4(x – 1)} + \frac{1}{2(x – 1)^2} + \frac{1}{4(x + 1)}$$

Similarly, all other partial fractions in the case of a repeated linear root may be found.

 Case III: Q(x) contains a higher order polynomial factor

Under this case, some factors of Q(x) might be quadratic, cubic or higher order polynomials in nature. Let us understand this method with respect to a quadratic term in the denominator –

Example – To find the partial fractions corresponding to \(\frac{x^3 – 2}{x^4 – 1}\):

First let us factorize the denominator – \(x^4 – 1 = (x^2 – 1)(x^2 + 1) = (x – 1)(x + 1)(x^2 + 1)\).
Now, let $$\frac{x^3 – 2}{(x – 1)(x + 1)(x^2 + 1)} = \frac{A}{(x – 1)} + \frac{B}{(x + 1)} + \frac{Cx + D}{(x^2 + 1)}$$ $$= \frac{A(x + 1)(x^2 + 1) + B(x – 1)(x^2 + 1) + (Cx + D)(x – 1)(x + 1)}{(x – 1)(x + 1)(x^2 + 1)}$$

Basically, we have chosen the numerator for the partial fraction involving the quadtratic term as a linear function in x: Cx + D (i.e. one degree less than the denominator).

To find A,B,C and D, now we will equate the coefficients of the numerators on both sides of the equation, since the fraction represented by them is the same. We then get –

x³ – 2 = A(x + 1)(x² + 1) + B(x – 1)(x² + 1) + (Cx + D)(x – 1)(x + 1)

Now we can proceed by equating the coefficients of equal powers of x on both sides; or once again proceed with direct substitution.

Note that on substituting x = 1 in the equation, we get –

1³ – 2 =A(1 + 1)(1² + 1) + B(1 – 1)(1² + 1) + (C.1 + D)(1 – 1)(1 + 1)
-1 = 4A + 0 + 0
A = -1/4

Similarly, on substituting x = -1;

(-1)³ – 2 =A(-1 + 1)((-1)² + 1) + B(-1 – 1)((-1)² + 1) + (C.(-1) + D)(-1 – 1)(-1 + 1)
-3 = 0 – 4B + 0
B = 3/4

Now we have no option but to compare the coefficients of same powers of x on both sides.

Coefficient of x³ on LHS  = Coefficient of x³ on RHS
1 = A + B + C
Use values of A and B to get –
C = 1/2

Constant term on LHS  = Constant term on RHS
1 = A – B – D
Use values of A and B to get –
D = 1

Our final result then can be written as -$$\frac{x^3 – 2}{(x – 1)(x + 1)(x^2 + 1)} = \frac{A}{(x – 1)} + \frac{B}{(x + 1)} + \frac{Cx + D}{(x^2 + 1)}$$ $$ = -\frac{1}{4(x – 1)} + \frac{3}{4(x + 1)} + \frac{x/2 + 1}{(x^2 + 1)}$$ $$ = -\frac{1}{4(x – 1)} + \frac{3}{4(x + 1)} + \frac{x + 2}{2(x^2 + 1)}$$

Similarly, all other partial fractions in the case of a higher polynomial function may be found.

Definite Integration

All of the integration fundamentals that you have studied so far have led up to this point, wherein now you can apply the integral evaluation techniques to more practical situations by incorporating the boundaries (or the limits) to which your integrand actually holds true.

This is that branch of integration which is of widespread use in Economics, Mathematics, Engineering, and many other disciplines. Its utilization in the task of finding the area under a curve finds great application in estimating the total revenues from related marginal functions, or the total growth from the growth rates trend graph etc. So let’s understand its working formula!

Given an integral \(\int f(x)dx\). We call the evaluation of such an integral as Indefinite Integration.

Formula

The solution of this integration is a resultant function in x plus some arbitrary constant. Let us represent the solution in this form –

\(\int f(x)dx = F(x) + c\)

In the method of definite integration, the integral actually has to evaluated in some domain of the variable x. Therefore, we represent it by \(\int_{x_1}^{x_2}\). What this actually means is that the integrand f(x) now will be bound over the values which the variable of integration i.e. x assumes here, which is x₁ to x₂. Then we can give the end result by –

\(\int_{x_1}^{x_2} f(x)dx = F(x_2) – F(x_a)\)
Here, x₂ – The Upper Limit
x₁ – The Lower Limit

Note that there is no arbitrary constant of integration in the solution of a definite integration. Also, you need to know beforehand the result of the indefinite integration of f(x), then use its value at the limits of integration to get this solution.

Properties

For a,b,c as real numbers –

\(\int_a^b f(x)dx = \int_a^b f(t)dt\) {Change of variable of integration}

⇒ \(\int_a^b f(x)dx = – \int_b^a f(x)dx\) {Reversal of Lower and Upper Limits}

⇒ \(\int_a^b f(x)dx = \int_a^c f(x)dx + \int_c^b f(x)dx, where a<c<b

⇒ \(\int_0^a f(x)dx = \int_0^a f(a – x)dx \)

⇒ If f(x) = f(a + x) {Periodic Function}; \(\int_0^{na} f(x)dx =  n\int_0^a f(x)dx\)

⇒ \(\int_{-a}^a f(x)dx = 2\int_{0}^a f(x)dx\) if f(-x) = f(x) {Even Function}
= 0 if f(-x) = -f(x) {Odd Function}

These properties are very useful for changing the limits, or simplifying the integral given in the problem. Thus, we must understand and memorize them well. Now let us look at some solved examples on what we have learned so far.

Solved Examples for You

Question: Solve $$ \int{\frac{1}{(x – 1)^2(x + 1)}}dx $$
Solution: We can evaluate this type of integrand easily by partial fractions method. In the example for Case 2) above, we have expressed exactly this integrand in terms of partial fractions. Now let us proceed further to find the complete solution.
$$ \int \frac{1}{(x – 1)^2(x + 1)} dx $$ $$ = \int[- \frac{1}{4(x – 1)} + \frac{1}{2(x – 1)^2} + \frac{1}{4(x + 1)}]dx $$ $$ \text{Using the standard formulae and the variable substitution method -} $$ $$ = -\frac{1}{4}ln|x – 1|  – \frac{1}{2(x – 1)} + \frac{1}{4}ln|x + 1| + c $$ $$ = \frac{1}{4}ln |\frac{x + 1}{x – 1}| – \frac{1}{2(x – 1)} + c $$

Question: Solve $$ \int{x^2lnx dx}$$
Solution: We’ll use integration by parts here with x² (algebraic) as the second function, and ln x (logarithmic) as the first function. Using the formula, we get –
$$ \int{x^2lnx dx}=ln x\int x^2 dx – \int[\frac{d}{dx}(ln x)\int x^2 dx]dx + c $$ $$ = ln x. \frac{x^3}{3} – \int[\frac{1}{x}.\frac{x^3}{3}]dx + c$$ $$ = \frac{x^3 ln x}{3} – \int{\frac{x^2}{3} dx}+c$$ $$ = \frac{x^3 ln x}{3} – \frac{x^3}{9} + c$$  

Question: Solve $$ \int_{-2}^{+2} x^2 dx $$
Solution: First, you must note that for this definite integral, the integrand x3 is an even function, that is –

(-x)² = (x)²

Thus, by the properties of such definite integrals we have – $$\int_{-2}^{+2} x^2 dx = 2\int_{0}^{2} x^2dx$$ $$ = 2[\frac{x^3}{3}]_0^2 $$ $$ = 2.[\frac{2^3}{3} – \frac{0^3}{3}] $$ $$ = \frac{16}{3}$$