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, replace all letters x by u to get 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 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 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 the variable \( x \), and the right is in terms of \( u \).

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Frequently Asked Questions

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

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

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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 \( u = x^5+2 \).

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What is the formula of integration by substitution?

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

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What are the examples 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 \( \displaystyle \frac{d}{dx}(x^4+10) = 4x^3 \)

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Practice Question

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


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