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:


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


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

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


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


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 $t \, dt = \frac{1}{6} \, du$.


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

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:

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


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


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 $x$:

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