## U-Substitution

Let us consider the impact of reversing the chain rule for differentiation when it comes to finding antiderivatives and indefinite integrals.

Recall that the chain rule states for differentiable functions $f(x)$ and $g(x)$, we have

$$\frac{d}{dx}[f(g(x))] = f'(g(x)) \cdot g'(x)$$

Expressing this in terms of integration, yields

$$\int f'(g(x)) \cdot g'(x) \, dx = f(g(x)) + C$$

Rephrased slightly, if $F(x)$ is an antiderivative of $f(x)$, then

$$\int f(g(x)) \cdot g'(x) \, dx = F(g(x)) + C$$

Of course to use this rule, we need to see to the right of the integral sign a composition with some expression $g(x)$ on the inside of that composition, along with the derivative of that $g(x)$ on the outside of that composition, as a factor of the entire expression to be integrated.

Suppose we denoted by $u$ the inside of the composition. That is to say, $u = g(x)$.

Differentiating both sides and then multiplying by the differential $dx$ then tells us that $du = g'(x) \, dx$. This let's us write the above integral in a different way, as

$$\int f(u) \, du$$

which may be easier to integrate.

Let us consider a few examples of this process, called $u$-substitution, to make the technique more clear:

#### Example

Find $\int \sqrt{3x+4} \, dx$

#### Solution:

Noting that we seek the integral of a composition, let $u = 3x+4$ so that it equals the inside of that composition.

Then note that $du = 3 \, dx$, which immediately tells us that $dx = \frac{1}{3} \, du$.

Replacing the expressions in terms of $x$ (including $dx$) with their corresponding expressions in terms of $u$ then gives

$$\int \sqrt{3x+4} \, dx = \int \sqrt{u} \cdot \frac{1}{3} \, du$$

Pulling out the factor of $\frac{1}{3}$ and rewriting the $\sqrt{u}$ as a rational power reveals an expression easy to integrate.

$$\begin{array}{rcl} \int \sqrt{u} \cdot \frac{1}{3} \, du &=& \frac{1}{3} \int \sqrt{u} \, du\\ &=& \frac{1}{3} \int u^{1/2} \, du\\ &=& \frac{1}{3} \cdot \frac{2}{3} u^{3/2} + C\\ &=& \frac{2}{9} u^{3/2} + C \end{array}$$

Finally, we would of course prefer to have our answer in terms of the original variable $x$, so let us appeal to $u = 3x+4$ one more time, as we conclude

$$\int \sqrt{3x+4} \, dx = \frac{2}{9} (3x+4)^{3/2} + C$$

#### Example

Find $\int t (5+3t^2)^8 \, dt$

#### Solution:

Again, we see a composition in the expression to be integrated. Let $u = 5 + 3t^2$, so it equals the inside of that composition.

Then note that $du = 6t \, dt$, so $t \, dt = \frac{1}{6} \, du$.

Hence,

$$\begin{array}{rcl} \int t (5 + 3t^2)^8 \, dt &=& \int (5+32^2)^8 \cdot (t \, dt)\\ &=& \int u^8 \cdot (\frac{1}{6} \, du)\\ &=& \frac{1}{6} \int u^8 \, du\\ &=& \frac{1}{6} \cdot \frac{1}{9} u^9 + C\\ &=& \frac{1}{54} u^9 + C \end{array}$$

Finally, to get our integral back in terms of the original variable $t$, we appeal to $u = 5 + 3t^2$ one more time to obtain:

$$\int t (5+3t^2)^8 \, dt = \frac{1}{54} (5 + 3t^2)^9 + C$$

#### Example

Find $\int x^2 \sqrt{1+x} \, dx$

#### Solution:

This time we may worry slightly at not seeing the derivative of the inside of the composition (or a simple multiple of it) sitting on the outside of the composition as a factor of the overall expression to be integrated.

Sometimes such a worry is well-founded. However, here there is an easy way to resolve the situation.

Again pick $u$ to be the inside of the composition seen. So $u = 1+x$. This can be used not only to give us a way to "substitute out" the $dx$, as $du = dx$ -- but also to replace other $x$ expressions as well. Note, if $u = 1+x$, then $x = u-1$.

Hence, we can rewrite the integral in the following way:

$$\begin{array}{rcl} \int x^2 \sqrt{1+x} \, dx &=& \int (u-1)^2 u^{1/2} \, du\\ &=& \int (u^2 - 2u + 1) u^{1/2} \, du\\ &=& \int u^{5/2} \, du -2 \int u^{3/2} \, du + \int u^{1/2} \, du\\ &=& \frac{2}{7} u^{7/2} - \frac{4}{5} u^{5/2} + \frac{2}{3} u^{3/2} + C \end{array}$$

Finally, using $u = 1+x$ one last time, we write the expression sought in terms of the original variable $x$:

$$\int x^2 \sqrt{1+x} \, dx = \frac{2}{7} (1+x)^{7/2} - \frac{4}{5} (1+x)^{5/2} + \frac{2}{3} (1+x)^{3/2} + C$$