Is there a chain rule for integration?

Solution 1:

The "chain rule" for integration is the integration by substitution.

$$\int_a^b f(\varphi(t)) \varphi'(t)\text{ d} t = \int_{\varphi(a)}^{\varphi(b)} f(x) \text{ d} x $$

So in your case we have $f(x) = x^5$ and $\varphi(t) = 2t+3$:

$$ \int (2t + 3)^5 \text{ d}t = \int {1 \over 2}\left((2t + 3)^5\cdot2\right) \text{ d}t = {1\over 2}\int x^5 \text{ d}x = {1\over 12} x^6 + C= {1\over 12} (2t+3)^6 + C$$

Solution 2:

If we know the integral of each of two functions, it does not follow that we can compute the integral of their composite from that information.

Example
$$ \int e^{-x}\;dx = -e^{-x} +C\\ \int x^2\;dx = \frac{x^3}{3} +C\\ $$ but $$ \int e^{-x^2}\;dx = \frac{\sqrt{\pi}}{2}\;\mathrm{erf}(x) + C $$ is not an elementary function.

Solution 3:

I'm guessing you're asking how to do the integral

$$\int (2x+3)^5 \, dx$$

I would use substitution:

$$u=2x+3 \\ du=2 \, dx$$

So your new integral is

$$\int \frac{u^5}2 \, du = \frac{u^6}{12} +C$$

Then you replace $u$ with the original $2x+3$ to get

$$\int \frac{u^5}2 \, du = \frac{u^6}{12} +C = \frac{(2x+3)^6}{12} +C$$

If you want to see how this relates to the chain rule, take the derivative of your answer, and it should get you the function "inside" the original integral.

$$F(x)=\frac{(2x+3)^6}{12} = f(g(x))$$ $$f(x)=\frac{x^6}{12} \, \, \, g(x)=2x+3 \\ f'(x)=\frac{x^5}2 \, \, \, g'(x)=2 \\$$

Using the chain rule we get

$$F'(x) = f'(g(x))g'(x) = f'(2x+3)g'(x) = \frac{(2x+3)^5}2 (2) = (2x+3)^5$$

Solution 4:

No.

Sorry for turning up late here, but I think the other (excellent) answers miss a key point. There is no direct, all-powerful equivalent of the differential chain rule in integration. The existence of the chain rule for differentiation is essentially what makes differentiation work for such a wide class of functions, because you can always reduce the complexity. The absence of an equivalent for integration is what makes integration such a world of technique and tricks.

The key point I speak of, therefore, is that hardly any functions can be integrated! Given a function of any complexity, the chances of its antiderivative being an elementary function are very small.

This is deeply contrary to the expectations you build when learning integration - but that's because the lessons are focusing on functions you can integrate, which fortunately overlap closely with the sorts of elementary functions you'd have learned at that stage: trig, exp, polynomials, inverses. They don't focus on the absence of techniques on non-integrable functions, because there's not much to say, and that leaves the impression that having an elementary antiderivative is the norm. And when you think about it, the key technique in integration is spotting how to turn what you've got into the result of a differentiation, so you can run it backwards.

Fortunately, many of the functions that are integrable are common and useful, so it's by no means a lost battle. And the mine of analytical tricks is pretty deep. And when that runs out, there are approximate and numerical methods - Taylor series, Simpsons Rule and the like, or, as we say nowadays "computers" - for solving anything definite. Or we just give the result a nice name (eg erf) and leave it at that.

Oblig. xkcd: https://xkcd.com/2117/

Solution 5:

The chain rule for integration is basically $u$-substitution.