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It's a combination of parts and substitution. You can start from the integral of an inverse function. With ##f(x) = t##, do a change of variables, followed by parts.gleem said:According to the author of this article, this integration formula is well-known but rarely taught. Do you know it?
$$\int f(x)dx = xf(x)-\int_{x_{0}}^{f(x)}f^{-1}(t)dt $$
where x0 is a constant and f-1(x) is the inverse function of f(x).
The integral of ##\log t## is ##t\log t - t##.fresh_42 said:Is there a typo somewhere? I get for ##f(x)=e^x##
\begin{align*}
\int e^x\,dx &\stackrel{?}{=} xe^x-\int_{x_{0}}^{e^x}\log(t)\,dt\\
&= xe^x-\left[t\log(t)-1\right]_{x_{0}}^{e^x}\\
&=xe^x-(e^x x-1)+(x_0\log(x_0)-1)\\
&=x_0\log(x_0)
&\neq e^{x}
\end{align*}
Where am I wrong?
You can also just take the derivative of the RHS, and verify that it is f(x). I note that I have never seen this before. It looks cute, but I wouldn't know how useful it is without playing with some examples myself.PeroK said:It's a combination of parts and substitution. You can start from the integral of an inverse function. With ##f(x) = t##, do a change of variables, followed by parts.
$$\int_{t_0}^{t_1}f^{-1}(t)dt = \int_{x_0}^{x_1}xf'(x)dx$$$$ =\bigg[xf(x)\bigg]_{x_0}^{x_1} - \int_{x_0}^{x_1}f(x)dx$$
This looks like one of those integration identities that might find use on some ubsurd integral challenges (which I enjoy), but probably not much use anywhere else. I would be interested to see if it does show up in someones research somewhere.PAllen said:It looks cute, but I wouldn't know how useful it is without playing with some examples myself.
It could be useful, but it doesn't have a lot of independent value. It's just the full substitution, followed by parts. In that sense, it's nothing new.Mondayman said:This looks like one of those integration identities that might find use on some ubsurd integral challenges (which I enjoy), but probably not much use anywhere else. I would be interested to see if it does show up in someones research somewhere.