# 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 $u$.

### 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.

### 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 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 $\displaystyle \frac{d}{dx}(x^4+10) = 4x^3$

## Practice Question

Take time to practise similar questions of the $u$-substitution integration method.