Integration by parts is a special technique of integration of two functions when they are multiplied. This method is also termed as partial integration. Another method to integrate a given function is integration by substitution method. These methods are used to make complicated integrations easy. Mathematically, integrating a product of two functions by parts is given as:

∫f(x).g(x)dx=f(x)∫g(x)dx−∫f′(x).(∫g(x)dx)dx

Integration By Parts Formula

If u and v are any two differentiable functions of a single variable x. Then, by the product rule of differentiation, we have;

d/dx(uv) = u(dv/dx) + v(du/dx)

By integrating both the sides, we get;

uv = ∫u(dv/dx)dx + ∫v(du/dx)dx

or

∫u(dv/dx)dx = uv-∫v(du/dx)dx  ………….(1)

Now let us consider,

u=f(x) and  dv/dx = g(x)

Thus, we can write now;

du/dx = f'(x) and v = ∫g(x) dx

Therefore, now equation 1 becomes;

∫f(x) g(x) dx = f(x)∫g(x) dx – ∫[∫g(x) dx] f'(x) dx

or

∫f(x) g(x) dx = f(x)∫g(x)dx – ∫[f'(x)∫g(x)dx]dx

This is the basic formula which is used to integrate products of two functions by parts.

If we consider f as the first function and g as the second function, then this formula may be pronounced as:
“The integral of the product of two functions = (first function) × (integral of the second function) – Integral of [(differential coefficient of the first function) × (integral of the second function)]”.

Also, read:

• Continuous Integration
• Integration Rules
• Methods Of Integration

ILATE Rule

Identify the function that comes first on the following list and select it as f(x).

ILATE stands for:

I: Inverse trigonometric functions : arctan x, arcsec x, arcsin x etc.

L: Logarithmic functions : ln x, log5(x), etc.

A: Algebraic functions.

T: Trigonometric functions, such as sin x, cos x, tan x etc.

E: Exponential functions.

Integration by parts uv formula

As derived above, integration by parts uv formula is:

\(\int du(\frac{dv}{dx})dx=uv-\int v(\frac{du}{dx})dx\)

Here, 

u = Function of u(x)

v = Function of v(x)

dv = Derivative of v(x)

du = Derivative of u(x)

Integration by parts with limits

In calculus, definite integrals are referred to as the integral with limits such as upper and lower limits. It is also possible to derive the formula of integration by parts with limits. Thus, the formula is:

\(\int_{a}^{b} du(\frac{dv}{dx})dx=[uv]_{a}^{b}-\int_{a}^{b} v(\frac{du}{dx})dx\)

Here,

a = Lower limit

b = Upper limit

Lets Work Out

Examples

Examples- Evaluate \(\int x.e^{x}dx\)

Solution- From ILATE theorem, f(x) = x, and g(x) = \(e^{2}\)

Thus using the formula for integration by parts, we have

\(\int f(x).g(x)dx = f(x)\int g(x)dx-\int f'(x).( \int g(x)dx )dx\)

\(\int x.e^{x}dx\) = \(x.\int e^{x}dx – \int 1. (\int e^{x}dx)dx\)

= \(x.e^{x} – e^{x} + c\)

Example- Evaluate \(\int \sqrt{x^{2}- a^{2}}\)

Solution- Choosing first function to be \(\sqrt{x^{2}- a^{2}}\) and second function to be 1.

\(\int \sqrt{x^{2}- a^{2}}\) = \(\sqrt{x^{2}- a^{2}}\int 1.dx – \int \frac{1}{2}.\frac{2x}{\sqrt{x^{2}- a^{2}}}.(\int 1.dx).dx\)

I = \(x.\sqrt{x^{2}- a^{2}} – \int \frac{x^{2}}{\sqrt{x^{2}- a^{2}}}.dx\)

Adding and subtracting a2 in the latter part of the integral we have

I = \(x.\sqrt{x^{2}- a^{2}} – \int \frac{x^{2}-a^{2}+a^{2}}{\sqrt{x^{2}- a^{2}}}.dx\)

I = \(x.\sqrt{x^{2}- a^{2}} – \int \frac{x^{2}-a^{2}}{\sqrt{x^{2}- a^{2}}}.dx – \int \frac{a^{2}}{\sqrt{x^{2}- a^{2}}}.dx\)

I = \( x.\sqrt{x^{2}- a^{2}}\) – I – \(a^{2} \int \frac{1}{\sqrt{x^{2}- a^{2}}}.dx\)

2I = \(x.\sqrt{x^{2}- a^{2}} – a^{2} \log \left | x + \sqrt{x^{2}- a^{2}} \right | + C\)

I = \(= \frac{x.\sqrt{x^{2}- a^{2}}}{2} – \frac{a^{2}}{2} \log \left | x + \sqrt{x^{2}- a^{2}} \right | + C_{1}\)

Example- Evaluate \(\int_{0}^{1}\arctan x .dx\)

Solution- Let

u = \(\arctan x\)                    dv = dx

\(du = \frac{1}{1+x^{2}}.dx\)             v = x

Integration by parts-

\(\int_{0}^{1}\arctan x .dx\) = \(= \left ( x\arctan x \right )_{0}^{1} – \int_{0}^{1}\frac{x}{1 + x^{2}}dx\)

= \( \left ( \frac{\pi}{4} – 0 \right ) – \left ( \frac{1}{2} \ln (1+ x^{2}) \right )_{0}^{1}\)

= \( \left ( \frac{\pi}{4} \right ) – \frac{1}{2} \ln 2\)

= \( \left ( \frac{\pi}{4} \right ) – \ln \sqrt{2}\)

Learn more about Integration, Integration by Substitution and many more. Register with BYJU’S today and get access to free material on various concepts.