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Let's say I wish to evaluate $$\int\frac{\sin(x)}{x}+\log(x)\cos(x)\ dx$$and$$\int\frac{xe^{2x}}{(1+2x)^2}\ dx$$

I can recognize at once the integrals as antiderivatives, or results of the product rule, quotient rule, power rule, etc.

For example in the first integral, $$\frac{d(\sin(x)\log(x)}{\ dx}=\frac{\sin(x)}{x}+\log(x)\cos(x)$$ thus $$\int\frac{\sin(x)}{x}+\log(x)\cos(x)\ dx= \sin(x)\log(x) +C$$

The same is true for the second integral by a multiplicative constant, $$\frac{\ d\big(\frac{e^{2x}}{4(1+2x)}\big)}{dx}=\frac{xe^{2x}}{(1+2x)^2}$$ thus $$\int\frac{xe^{2x}}{(1+2x)^2}\ dx=\frac{e^{2x}}{4(1+2x)}+C$$

My question is, does there exist an elementary path for these integrals, by that I mean can these integrals be explicitly evaluated in an easy way, if there exists a simple product rule, quotient rule, power rule, etc derivation such as the ones above? I tried the first one for a long time but couldn't find a way to combine the fractions and make a substitution.

Thanks in advance!

coreyman317
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    One of my professors said: "computing derivatives is just following an algorithm, but computing integrals is an art". I guess the closest to an "inverse product rule" etc is integration by parts. But in general it is non-trivial to detect that a product rule was used. – M. Winter Jul 09 '18 at 07:23
  • I think the situation is even worse for the quotient-rule. The Risch-algorithm should find such antiderivates, but I do not think it gives a straightford method. – Peter Jul 09 '18 at 07:26
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    The standard approach to showing an integral cannot be written with elementary functions is called Liouville's Criterion. For example, see my answer here for a reference that proves the antiderivative of $\sin(x)/x$ to be inexpressible with such terms. – Benjamin Dickman Jul 09 '18 at 07:26
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    Usually there is no generalized way, that's the reason we need an integral table to help memorize commonly used intergals, e.g, the intergral of combination of fundamental functions like polyniomial, exponential, triangular, log, etc. – J. Yu Jul 09 '18 at 07:27

2 Answers2

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As you just have shown that $$\int\big(\frac{\sin(x)}{x}+\log(x)\cos(x)\big)\ dx= \sin(x)\log(x) +C$$ and as it is known that $$\int\frac{\sin(x)}{x}\ dx$$ has no explicite antiderivative (in terms of "usual" functions - it well can be just being defined such, cf. $Si(x)$), it becomes obvious that $$\int\log(x)\cos(x)\ dx$$ will not have either.

--- rk

Dr. Richard Klitzing
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Your first integral can be evaluated by separating the two terms and integrating the first term by parts, like this: $$\int\Big(\frac{\sin x}{x}+\log x\cos x\Big)\,dx=\int\frac{\sin x}{x}\,dx+\int\log x\cos x\,dx=\\=\Big(\log x\sin x-\int\log x\cos x\,dx\Big)+\int\log x\cos x\,dx=\log x\sin x +C,$$ where the $\int F(x)g'(x)\,dx$-term produced by the partial integration cancels out the other integral.

Your second integral is much more complicated and would be harder to evaluate using general methods, although it should still be possible.

apanpapan3
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