# Integration by parts – formula, proof, derivation, examples and FAQs

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

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.

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

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-1x.

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-1x.dx = ∫tan-1x.1.dx

= tan-1x.∫1.dx – ∫((tan-1x)’.∫ 1.dx).dx

= tan-1x. x – ∫(1/(1 + x2).x).dx

= x. tan-1x – ∫ 2x/(2(1 + x2)).dx

= x. tan-1x – ½.log(1 + x2) + 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.

∫ ex(f(x) + f'(x)).dx = exf(x) + C
∫√(x2 + a2).dx = ½ . x.√(x2 + a2)+ a2/2. log|x + √(x2 + a2)| C
∫√(x2 – a2).dx =½ . x.√(x2 – a2) – a2/2. log|x +√(x2 – a2) | C
∫√(a2 – x2).dx = ½ . x.√(a2 – x2) + a2/2. sin-1 x/a + C
☛ Also Check: https://www.cuemath.com/calculus/integration-by-parts/
Example

Find ∫xe-x dx

Integrating by parts (with v = x and du/dx = e-x), we get:

-xe-x – ∫-e-x dx (since ∫e-x dx = -e-x)

= -xe-x – e-x + constant

We can also sometimes use integration by parts when we want to integrate a function that cannot be split into the product of two things. The trick we use in such circumstances is to multiply by 1 and take du/dx = 1.

EXAMPLE

Find ∫ ln x dx

To integrate this, we use a trick, rewrite the integrand (the expression we are integrating) as 1.lnx . We then let v = ln x and du/dx = 1 .

Hence ∫ ln x dx = x ln x – ∫ x (1/x) dx
= x lnx – ∫ dx
= x lnx – x + constant

Solved Examples On Integration By Parts
Example 1: Find the integral of x2ex by using the integration by parts formula.

Solution:

Using LIATE, u = x2 and dv = ex dx.

Then, du = 2x dx, v = ∫ ex dx = ex.

Using one of the integration by parts formulas,

∫ u dv = uv – ∫ v du

∫ x2 ex dx = x2 ex – ∫ ex (2x) dx

= x2 ex – 2 ∫ x ex dx

Applying integration by parts formula again to evaluate ∫ x ex dx,

∫ x2 ex dx = x2 ex – 2 (x ex – ∫ ex dx) = x2 ex – 2 x ex + 2 ex + C

= ex (x2- 2 x + 2)+ C

Answer: ∫ x2 ex dx = = ex (x2- 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) dxU

= ln x (x2/2) – ∫ (1/x) (x2/2) dx

= (x2 ln x)/2 – (1/2) ∫ x dx

= (x2 ln x) / 2 – (1/2) (x2/2) + C

= (x2 ln x) /2 – (x2 / 4) + C

=(x2/4)(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 = x2/2

Using one of the formulas of integration by parts,

∫ u dv = uv – ∫ v du

∫ x ln x dx = ln x (x2/2) – ∫ (x2/2) (1/x) dx

= (x2 ln x)/2 – (1/2) ∫ x dx

= (x2 ln x) / 2 – (1/2) (x2/2) + C

= (x2 ln x) /2 – (x2 / 4) + C

= (x2/4)(2 ln x -1) + C

Have questions on basic mathematical concepts?
Become a problem-solving champ using logic, not rules. Learn the why behind math with our certified experts
Book a Free Trial Class

Practice Questions on Integration by Parts
Q 1. Find ∫ sin-1x. dx using integration by parts.

x sin-1x -√(x2 -1 )+ C.
x sin, -1, x -√(x, 2, -1, , )+ C.

x sin-1x + √(1-x2) + C.
x sin, -1, x + √(1-x, 2, ) + C.

x sin-1x -√(1-x2) + C.
x sin, -1, x -√(1-x, 2, ) + C.

x sin-1x -√(x2 -1 ) + C.
x sin, -1, x -√(x, 2, -1, , ) + C.
Q 2. For which of the following integral expressions we need to apply Integration by Parts formula.

∫ log sinx dx
∫ log sinx dx

∫ x ex . dx
∫ x e, x , . dx

∫ cosec2 x. dx
∫ cosec, 2 , x. dx

∫ tan x. dx
∫ tan x. dx

FAQs on Integration by Parts
What is Integration by Parts?

What Is Integration By Parts Formula?
How To Derive Integration By Parts Formula?
Why do we Use Integration by Parts Formula?
What are the Different Techniques of Integration in addition to Integration by Parts?
How to Know When to Use Integration by Parts?
Which of the Function Should be Made as ‘U’ in Integration by Parts?
What is the Difference between Integration by Parts and Substitution?
How to Apply Limits in Integration by Parts?
What is the Application of Integration by Parts?
What Are the Applications of the Integration By Parts Formula?
How To Know When To Use the Integration By Parts Formula?

Integration by parts
Use Uv Method To Find Integration By Parts

Calculus
Differentiation and Integration Formulas
Differential Equations
Integration Formulas

day & time:

Book A Free Class

YOU CAN DO IT

### Oluwamuyide Peter

On the 4th of November I officially became a member of the exclusive 1st student with distinction after five years of no such record, in the history of The Polytechnic Ibadan, Faculty of Engineering to graduate with distinction as a DPP students since its establishment in 2011. My unrelenting power to solve problems, have made me to create a platform where student can get valid information anywhere, anyplace at anytime Evolving education world wide 🌎

Check Also
Close