Existence of elementary integral because of simple product rule?

Question

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!

Answer 1

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.

Answer 2

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