Easy Method of Integration by Substitution: U-Substitution

The substitution method of integration is useful when an integral contains some function and its derivative. Set a replacement letter, say u mostly (sometimes w) and in this case, and replace all letters x by u for getting the integration done easier. Once integrate the u-integral in terms of u, replace x-written expressions to get the proper solutions.
Though it is now always easy to identify which part of the integral is to be substituted, it is vital to determine \( u= \cdots \) to perform the integration using the \(u\)-substitution.

$$ \displaystyle \int{\frac{2x}{\sqrt{x^2-4}}}dx = \int{\frac{\frac{d}{dx}(x^2-4)}{\sqrt{x^2-4}}}dx $$

This method is also called the \( u \)-substitution method as the derivative is substituted as \( u \).

$$ u = x^2-4 $$

The next step is to differentiate this \(u\)-substitution in terms of \(x\).

$$ \displaystyle \frac{du}{dx} = 2x $$

Rearrange this expression by letting \(dx\) the subject.

$$ \displaystyle dx = \frac{du}{2x} $$

Now, the initial integral expression becomes:

$$ \displaystyle \int{\frac{2x}{\sqrt{x^2-4}}}dx = \displaystyle \int{\frac{2x}{\sqrt{x^2-4}}} \frac{du}{2x} $$

You can cancel \( 2x \) at the fraction.

$$ \begin{align} \displaystyle \int{\frac{2x}{\sqrt{x^2-4}}}dx &= \int{\frac{2x}{\sqrt{x^2-4}}} \frac{du}{2x} \\ &= \int{\frac{1}{\sqrt{x^2-4}}}du \end{align} $$

Replace \( x^2-4 \) by \( u \).

$$ \begin{align} \displaystyle \int{\frac{2x}{\sqrt{x^2-4}}}dx &= \int{\frac{2x}{\sqrt{x^2-4}}} \frac{du}{2x} \\ &= \int{\frac{1}{\sqrt{x^2-4}}}du \\ &= \int{\frac{1}{\sqrt{u}}}du \end{align} $$

This time, the \(x\)-integration ,\( \displaystyle \int{\frac{2x}{\sqrt{x^2-4}}}dx \) , becomes \(u\)-integration, \( \displaystyle \int{\frac{1}{\sqrt{u}}}du \), which is lot easier to solve. Perform the integration in terms of \( u \).

$$ \begin{align} \displaystyle \int{\frac{2x}{\sqrt{x^2-4}}}dx &= \int{\frac{2x}{\sqrt{x^2-4}}} \frac{du}{2x} \\ &= \int{\frac{1}{\sqrt{x^2-4}}}du \\ &= \int{\frac{1}{\sqrt{u}}}du \\ &= \int{u^{-\frac{1}{2}}}du \\ &= u^{\frac{1}{2}} + C \\ &= \sqrt{u} + C \end{align} $$

Almost completed, but don’t forget to replace the \(u\)-substitution back to \(x\)-expressions as the original question was regarding \(x \), not \(u\).

$$ \begin{align} \displaystyle \int{\frac{2x}{\sqrt{x^2-4}}}dx &= \int{\frac{2x}{\sqrt{x^2-4}}} \frac{du}{2x} \\ &= \int{\frac{1}{\sqrt{x^2-4}}}du \\ &= \int{\frac{1}{\sqrt{u}}}du \\ &= \int{u^{-\frac{1}{2}}}du \\ &= u^{\frac{1}{2}} + C \\ &= \sqrt{u} + C \\ &= \sqrt{x^2 – 4} +C \end{align} $$

Finally, we reached to our solution of the integration using \( u \)-substitution method.

$$ \begin{align} \therefore \displaystyle \int{\frac{2x}{\sqrt{x^2-4}}}dx &= \sqrt{x^2 – 4} +C \end{align} $$

Note that the integral on the left is expressed in terms of the variable \( x \), and the right is in terms of \( u \).

Frequently Asked Questions

How do we know when we can use u-substitution method of integration?

The substitution method of integration or \(u\)-substitution is used when an integral contains some function and its derivative.

How do we know which part of the integral is substituted?

When the difference of powers of two parts is 1, then take the higher part as \( u \) as the derivative of the higher power part will be the lower power part.
For example, \( \displaystyle \int{5x^4 (x^5+2)^3}dx \) has \( \displaystyle \frac{d}{dx}(x^5+2) = 5x^4 \), so it is determined \( u = x^5+2 \).

What is the formula of integration by substitution?

\( \displaystyle \int{f\left[g(x)\right]g^{\prime}(x)}dx = \int{f(u)}du \), where \( u=g(x) \)

What are the exmples of using \( u \)-substitution?

\( \displaystyle \int{\frac{3x^2}{(x^3+4)^5}}dx \leadsto u = x^3+4 \) as \( \displaystyle \frac{d}{dx}(x^3+4) = 3x^2 \)

\( \displaystyle \int{\frac{4x^3 + 6x}{(x^4+3x^2)^5}}dx \leadsto u = x^4+3x^2 \) as \( \displaystyle \frac{d}{dx}(x^4+3x^2) = 4x^3 + 6x \)

\( \displaystyle \int{4x^3(x^4+10)^5}dx \leadsto u = x^4+10 \) as \( \displaystyle \frac{d}{dx}(x^4+10) = 4x^3 \)

Practice Question

Take your time to practise similar questions of the \(u\)-substitution method of integration.

 



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