Integration by Parts
The idea of integration by parts was proposed in 1715 by Brook Taylor, who also proposed the famous Taylor's Theorem. Generally, integrals are calculated for functions for which differentiation formulas exist. Here Integration by parts is an additional technique used to find the integration of the product of functions and it is also referred to as partial integration. It changes the integration of the product of functions into integrals for which a solution can be easily computed.
Some of the inverse trigonometric functions and logarithmic functions do not have integral formulas, and here we can make use of integration by parts formula. Here we shall check the proof, the graphical representation, applications, and examples of integration by parts.
What is Integration by Parts?
Integration by parts is used to integrate the product of two or more functions. The two functions to be integrated f(x) and g(x) are of the form \(\int\)f(x).g(x). Thus, it can be called a product rule of integration. Among the two functions, the first function f(x) is selected such that its derivative formula exists, and the second function g(x) is chosen such that an integral of such a function exists.
\(\int f(x).g(x).dx = f(x) \int g(x).dx  \int (f'(x) \int g(x).dx).dx + C \)
The integration of (First Function x Second Function) = (First Function) x (Integration of Second Function)  Integration of (Differentiation of First Function x Integration of Second Function).
In the integration by parts, the formula is split into two parts and we can observe the derivative of the first function f(x) in the second part, and the integral of the second function g(x) in both the parts. For simplicity, these functions are often represented as 'u' and 'v' respectively. The integration of uv formula using the notation of 'u' and 'v' is:
∫ u dv = uv  ∫ v du.
Integration By Parts Formula
The integration by parts formula is used to find the integral of the product of two different types of functions such as logarithmic, inverse trigonometric, algebraic, trigonometric, and exponential functions. The integration by parts formula is used to find the integral of a product. In the product rule of differentiation where we differentiate a product uv, u(x), and v(x) can be chosen in any order. But while using the integration by parts formula, for choosing the first function u(x), we have to see which of the following function comes first in the following order and then assume it as u.
 Logarithmic (L)
 Inverse trigonometric (I)
 Algebraic (A)
 Trigonometric (T)
 Exponential (E)
This can be remembered using the rule LIATE. Note that this order can be ILATE as well. For example, if we have to find ∫ x ln x dx (where x is an algebraic function and ln is a logarithmic function), we will choose ln x to be u(x) as in LIATE, the logarithmic function appears before the algebraic function. The integration by parts formula is defined in two ways. We can use either of them to integrate the product of two functions.
Integration By Parts Formula Derivation
The proof of integration by parts can be obtained from the formula of the derivative of the product of two functions. For the two functions f(x) and g(x), the derivative of the product of these two functions is equal to the sum of the derivatives of the first functions multiplied with the second function, and the derivative of the second function multiplied by the first function.
Let us derive the integration by parts formula using the product rule of differentiation. Consider two functions u and v. Let their product be y. i.e., y = uv. Applying the product rule of differentiation, we get
d/dx (uv) = u (dv/dx) + v (du/dx)
We will rearrange the terms here.
u (dv/dx) = d/dx (uv)  v (du/dx)
Integrating on both sides with respect to x,
∫ u (dv/dx) (dx) = ∫ d/dx (uv) dx  ∫ v (du/dx) dx
By cancelling the terms,
∫ u dv = uv  ∫ v du
Hence the integration by parts formula is derived.
Visualizing Integration by Parts
Consider a parametric curve (x, y) = (f(θ), g(θ)). Let us consider this curve to be integrable and a onetoone function. The integration by parts represents the area of the blue region from the below curve. Let us first consider the areas of the blue region and the yellow regions distinctly.
Consider the curve along the yaxis we have the function x(y) and across the limits [\(y_2, y_1\)]. Also we can consider the curve along the xaxis and have the function y(x) across the limits [\(x_2, x_1\)].
Area of the blue region = \(\int^{y_2}_{y_1} x(y).dy \)
Area of the yellow region = \(\int^{x_2}_{x_1} y(x).dx\)
The total area of these two regions is equal to the area of the larger rectangle minus the area of the smaller rectangle.
\(\int^{y_2}_{y_1} x(y).dy \) + \(\int^{x_2}_{x_1} y(x).dx\) = \([x.y(x)]^{x_2}_{x_1}\)
Without the definite integrals it can be written as.
∫ y.dx+ ∫ x.dy = xy
∫x.dy = xy  ∫ y.dx
Further this can be modified to obtain the integration by parts formula.
∫f(x).g(x).dx = f(x).∫ g(x).dx  ∫(f'(x) .∫ g(x).dx).dx
Applications of Integration by Parts
The application of this formula for integration by parts is for functions or expressions for which the formulas of integration do not exist. Here we try to include this formula of integration by parts and try to derive the integral. For logarithmic functions and for inverse trigonometry functions there are no integral answers. Let us try to solve and find the integration of log x and tan^{1}x.
Integration of Logarithmic Function
∫ logx.dx = ∫ logx.1.dx
= logx. ∫1.dx  ∫ ((logx)'.∫ 1.dx).dx
=logx.x ∫ (1/x .x).dx
=xlogx  ∫ 1.dx
=x logx  x + C
Integration of Inverse Trigonometric Function
∫ tan^{1}x.dx = ∫tan^{1}x.1.dx
= tan^{1}x.∫1.dx  ∫((tan^{1}x)'.∫ 1.dx).dx
= tan^{1}x. x  ∫(1/(1 + x^{2}).x).dx
= x. tan^{1}x  ∫ 2x/(2(1 + x^{2})).dx
= x. tan^{1}x  ½.log(1 + x^{2}) + C
Formulas Related to Integration by Parts
The following formulas have been derived from the integration by parts formula and are helpful in the process of integrations of various algebraic expressions.
 ∫ e^{x}(f(x) + f'(x)).dx = e^{x}f(x) + C
 ∫√(x^{2} + a^{2}).dx = ½ . x.√(x^{2} + a^{2})+ a^{2}/2. logx + √(x^{2} + a^{2}) C
 ∫√(x^{2}  a^{2}).dx =½ . x.√(x^{2}  a^{2})  a^{2}/2. logx +√(x^{2}  a^{2})  C
 ∫√(a^{2}  x^{2}).dx = ½ . x.√(a^{2}  x^{2}) + a^{2}/2. sin^{1 }x/a + C
☛ Also Check:
Solved Examples on Integration By Parts

Example 1: Find the integral of x^{2}e^{x} by using the integration by parts formula.
Solution:
Using LIATE, u = x^{2} and dv = e^{x} dx.
Then, du = 2x dx, v = ∫ e^{x} dx = e^{x}.
Using one of the integration by parts formulas,
∫ u dv = uv  ∫ v du
∫ x^{2} e^{x} dx = x^{2} e^{x}  ∫ e^{x} (2x) dx
= x^{2} e^{x}  2 ∫ x e^{x} dx
Applying integration by parts formula again to evaluate ∫ x e^{x} dx,
∫ x^{2} e^{x} dx = x^{2} e^{x}  2 (x e^{x}  ∫ e^{x} dx) = x^{2} e^{x}  2 x e^{x} + 2 e^{x} + C
= e^{x} (x^{2} 2 x + 2)+ C
Answer: ∫ x^{2} e^{x} dx = = e^{x} (x^{2} 2 x + 2)+ C

Example 2: Find the integral of x sin2x, by using integration by parts formula.
Solution:
To find the integration of the given expression we use the integration by parts formula: ∫ uv.dx = u∫ v.dx ∫( u' ∫ v.dx).dx
Here u = x, and v = Sin2x
∫x sin2x. dx
=x∫sin2xdx  d/dx. x.∫ sin2xdx. dx
=x. cos2x/2  ∫(1.cos2x/2). dx
=cos2x/2. dx + 1/2 cos2xdx
=xcos2x/2 + sin2x/4 + C
Answer: Thus ∫x sin2x dx = x cos2x/2 +sin 2x/4+ C

Example 3: Evaluate the integral ∫ x ln x dx using integration by parts.
Solution:
First Method:
Using LIATE, u = ln x and v = x.
Using one of the formulas of integration by parts,
∫ uv dx = u ∫ v dx  ∫ (u' ∫ v dx) dx
∫ x ln x dx = ln x ∫ x dx  ∫ (1/x) (∫ x dx) dx
= ln x (x^{2}/2)  ∫ (1/x) (x^{2}/2) dx
= (x^{2} ln x)/2  (1/2) ∫ x dx
= (x^{2} ln x) / 2  (1/2) (x^{2}/2) + C
= (x^{2} ln x) /2  (x^{2} / 4) + C
=(x^{2}/4)(2 ln x 1) + C
Second Method:
Using LIATE, u = ln x and dv = x dx.
Then du = (1/x) dx and v = ∫ x dx = x^{2}/2
Using one of the formulas of integration by parts,
∫ u dv = uv  ∫ v du
∫ x ln x dx = ln x (x^{2}/2)  ∫ (x^{2}/2) (1/x) dx
= (x^{2} ln x)/2  (1/2) ∫ x dx
= (x^{2} ln x) / 2  (1/2) (x^{2}/2) + C
= (x^{2} ln x) /2  (x^{2} / 4) + C
= (x^{2}/4)(2 ln x 1) + C
Answer: By both the methods, ∫ x ln x dx =(x^{2}/4)(2 ln x 1) + C
FAQs on Integration by Parts
What is Integration by Parts?
The integration by parts is the integration of the product of two functions. The two functions are generally represented as f(x) and g(x). Among the two functions, the first function f(x) is selected such that its derivative formula exists, and the second function g(x) is chosen such that an integral of such a function exists.
∫ f(x).g(x).dx = f(x) ∫ g(x).dx  ∫(f'(x) ∫g(x).dx).dx + C
What Is Integration By Parts Formula?
The integration by parts formula is a formula used to find the integral of the product of two different types of functions. The popular integration by parts formula is,
∫ u dv = uv  ∫ v du
Here, the first function 'u' should be chosen according to LIATE (Logarithmic (L), Inverse trigonometric (I), Algebraic (A), Trigonometric (T), Exponential (E)).
How To Derive Integration By Parts Formula?
Using the product rule of differentiation,
d/dx (uv) = u (dv/dx) + v (du/dx)
u (dv/dx) = d/dx (uv)  v (du/dx)
Taking integral on both sides,
∫ u (dv/dx) (dx) = ∫ d/dx (uv) dx  ∫ v(du/dx) dx
This gives,
∫ u dv = uv  ∫ v du
Thus, the integration by parts formula is derived.
Why do we Use Integration by Parts Formula?
The formula of integration of parts is used since the normal form of integration is not possible. Integration is generally possible for functions for which the derivative formula is available. Expressions such as logarithmic functions, inverse trigonometric functions cannot be integrated easily and hence the integrals are found using integration by parts formula.
What are the Different Techniques of Integration in addition to Integration by Parts?
The three different techniques used for integration are as follows.
(a) Integration by Substitution
(b) Integration by Partial Fractions
(c) Integration by Parts
How to Know When to Use Integration by Parts?
The integration by parts is used when the simple process of integration is not possible. If there are two functions and a product between them, we can take the integration between parts formula. Also for a single function, we can take 1 as the other functions and find the integrals using integration by parts. For example, we can integrate Sin^{1}x, Logx, xCosx, using this formula.
Which of the Function Should be Made as 'U' in Integration by Parts?
The formula for integration by parts is \(\int uv.dx = u\int v.dx  \int( u'\int v.dx).dx\). Here the function 'u' is chosen such that the derivative formula of this function can be calculated.
What is the Difference between Integration by Parts and Substitution?
The integration of parts can be used for finding the integrals of the product of two functions, f(x).g(x). The integration by substitution can be calculated for functions having subfunctions, f(g(x)). The integration by parts can be used for functions such as xcosx, e^{x}tanx, xe^{x}. And the integration by substitution can be used for functions such as sin(logx). \(\sqrt{tanx}\), cosec^{2}(5^{x}).
How to Apply Limits in Integration by Parts?
The limits for integrations by parts can be applied similar to the definite integrals. Applying the lower limit 'a', and the upper limit 'b' to integration by parts, we have \(\int^b_a uv.dx =[ u\int v.dx  \int( u'\int v.dx).dx]^b_a\)
What is the Application of Integration by Parts?
The application of this formula for integration by parts is for functions or expressions for which the derivative do not exist, and which cannot be integrated by the simple process of integration. Here we try to use the formula of integration by parts and try to find the integral of the product of two or more functions. We can apply this formula for logarithmic functions and for inverse trigonometric functions which cannot be integrated using the simple process of integration.
What Are the Applications of the Integration By Parts Formula?
The integration by parts formula is used to find the integral of the product of two different types of functions. Also, this formula is used to find the integral of various functions such as sin^{1}x, ln x, etc by assuming the second function as 1.
How To Know When To Use the Integration By Parts Formula?
When we come across an integral of the product of two functions, then we have to apply the integration formula. Sometimes, we use the integration by parts formula when there is a single function also such as ln x, sin^{1}x, tan^{1}x, etc.