Integration by Parts

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:


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


∫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


∫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 product 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:


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 :sinx, cosx , tan x etc .

E: Exponential functions.

Lets Work Out


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.

Leave a Comment

Your email address will not be published. Required fields are marked *