TITLE: Why is integration so much harder than differentiation?
QUESTION [129 upvotes]: If a function is a combination of other functions whose derivatives are known via composition, addition, etc., the derivative can be calculated using the chain rule and the like. But even the product of integrals can't be expressed in general in terms of the integral of the products, and forget about composition! Why is this?

REPLY [34 votes]: I guess the OP asks about the symbolic integration. Other answers already dealt with the numeric case where integration is easy and differentiation is hard.
If you recall the definition of the differentiation, you can see it's just a subtraction and division by a constant. Even if you can't do any algebraic changes, it won't get any more complex than that. But usually you can do many simplifications due to the zero limit, as many terms fall out as being too small. From this definition it can be shown that if you know the derivative of $f(x)$ and $g(x)$, then you can use these derivatives to express the derivative of $f(x) \pm g(x)$, $f(x)g(x)$ and $f(g(x))$. 
This makes symbolic differentiation easy as you just need to apply the rules recursively.
Now about integration. Integration is basically an infinite sum of small quantities. So if you see an $\int f(x) \, dx$. You can imagine it as an infinite sum of $(f_1 + f_2 + ...) \, dx$ where $f_i$ are consecutive values of the function. 
This means if you need to calculate integral of $\int (a f(x) + b g(x)) \,d x$. Then you can imagine the sum $((af_1 + bg_1) + (af_2 + bg_2) + ...) \,d x$. Using the associativity and distributivity, you can transform this into: $a(f_1 + f_2 +...)\,d x + b(g_1 + g_2 + ...)\,d x$.
So this means $\int (a f(x) + b g(x)) \, d x = a \int f(x) \,d x + b \int g(x) \, dx$.
But if you have $\int f(x) g(x) \, d x$, you have the sum $(f_1 g_1 + f_2 g_2 + ...) \,d x$. From which you cannot factor out the sum of $f$s and $g$s. This means there is no recursive rule for multiplication. 
Same goes for $\int f(g(x)) \,d x$. You cannot extract anything from the sum $(f(g_1) + f(g_2) + ...) \,d x$ in general.
So far, only linearity is the useful property. What about the  analogues of the Differentiation rules? We have the product rule:
$$\frac{d f(x)g(x) }{\, d x} = f(x) \frac{d g(x)}{\, d x} + g(x) \frac{d f(x)}{\, d x}.$$ Integrating both sides and rearranging the terms, we get the well-known integral by parts formula:
$$\int f(x) \frac{d g(x)}{\, d x} \, d x = f(x)g(x) - \int g(x) \frac{d f(x)}{\, d x} \, d x.$$
But this formula is only useful if $\frac{d f(x)}{dx} \int g(x) \, d x$ or 
$\frac{d g(x)}{dx} \int f(x) \, d x$ is easier to integrate than $f(x)g(x)$.
And it's often hard to see when this rule is useful. For example, when you try to integrate $\mathrm{ln}(x)$, it's not obvious to see that it's $1 \mathrm{ln}(x)$. The integral of $1$ is $x$ and the derivative of $\mathrm{ln}(x)$ is $\frac{1}{x}$, which lead to a very simple integral of $x\frac{1}{x} = 1$, whose integral is again $x$.
Another well-known differential rule is the chain rule $$\frac{d f(g(x))}{\, d x} = \frac{d f(g(x))}{d g(x)} \frac{d g(x)}{\, d x}.$$
Integrating both sides, you get the reverse chain rule:
$$f(g(x)) = \int \frac{d f(g(x))}{d g(x)} \frac{d g(x)}{\, d x} \, d x.$$
But again it's hard to see when it is useful. For example what about the integration of $\frac{x}{\sqrt{x^2 + c}}$? Is it obvious to  you that $\frac{x}{\sqrt{x^2 + c}} = 2x \frac{1}{2\sqrt{x^2 + c}}$ and this is the derivative of $\sqrt{x^2 + c}$? I guess not, unless someone showed you the trick.
For differentiation, you can mechanically apply the rules. For integration, you need to recognize patterns and even need to introduce cancellations to bring the expression into the desired form and this requires lot of practice and intuition.
For example how would you integrate $\sqrt{x^2 + 1}$? 
First you turn it into a fraction:
$$\frac{x^2 + 1}{\sqrt{x^2+1}}$$
Then multiply and divide by 2:
$$\frac{2x^2 + 2}{2\sqrt{x^2+1}}$$
Separate the terms like this:
$$\frac{1}{2}\left(\frac{1}{\sqrt{x^2+1}}+\frac{x^2+1}{\sqrt{x^2+1}}+\frac{x^2}{\sqrt{x^2+1}} \right)$$
Play with 2nd and 3rd term:
$$\frac{1}{2}
\left(
\frac{1}{\sqrt{x^2+1}}+
1\sqrt{x^2+1}+
x2x\frac{1}{2\sqrt{x^2+1}} 
\right)$$
Now you can see the first bracketed term is the derivative of $\mathrm{arsinh(x)}$. The second and third term is the derivative of the $x\sqrt{x^2+1}$. Thus the integral will be:
$$\frac{\mathrm{arsinh}(x)}{2} + \frac{x\sqrt{x^2+1}}{2} + C$$
Were these transformations obvious to you? Probably not. That's why differentiation is just a mechanic while integration is an art.