# Examples of Integration by Parts With $v^\prime(x) = 1$

It is a well-known fact that the best way to tackle an integral such as $$\int \ln x \ \mathrm{d}x\qquad \mathrm{or}\qquad \int \arctan x \ \mathrm{d}x$$ is to use integration by parts, defining $$u(x) = \ln x$$ in the first case, $$u(x) = \arctan x$$ in the second, and $$v^\prime(x) = 1$$ in both cases.

It occurred to me that I have very rarely seen other integrals where setting $$v^\prime(x) = 1$$ is key to obtaining a solution via integration by parts, so I'd be very interested to see some more examples where this is the case. The more exotic the better!

This trick is helpful for integrating any inverse trigonometric function (that includes the inverse of hyperbolic trig functions). For example, $$\int (\sin^{-1} x)dx=\int (\sin^{-1}x)\cdot (1) dx=(\sin^{-1}x)\cdot x-\int \frac{x}{\sqrt{1-x^2}} dx =(\sin^{-1} x) \cdot x + (1-x^2)^{\frac{1}{2}}+C.$$

As another example, for $$x\geq 1,$$ $$\int (\sec^{-1} x) dx = \int\ (\sec^{-1} x) \cdot (1) dx = (\sec^{-1} x) \cdot x - \int \frac{x}{x\sqrt{x^2-1}} dx =(\sec^{-1} x) \cdot x-\int \frac{1}{\sqrt{x^2-1}} dx = (\sec^{-1} x) \cdot x-\ln(x+\sqrt{x^2-1})+C.$$

In fact the $$\ln$$ function is related to the inverse hyperbolic trig functions. For example, if $$\cosh x=y$$ then $$x=\ln(y\pm \sqrt{y^2-1}),$$ which shows that $$\cosh^{-1}(y)=\ln(y+\sqrt{y^2-1})$$ suggesting that this method is useful for $$\log$$ functions for essentially the same reason it is useful for inverse trigonometric functions.

This idea actually works for any sufficiently nice inverse function: if $$F' = f$$, we have $$\int f^{-1}(y) \, dy = y f^{-1}(y) - ( F \circ f^{-1} )(y) + C .$$ This result is actually true for quite general $$f^{-1}$$, but for differentiable ones, integration by parts works: firstly, $$\int 1 \cdot f^{-1}(y) \, dy = y \cdot f^{-1}(y) - \int y \cdot (f^{-1})'(y) \, dy ,$$ and then $$y = f(f^{-1}(y))$$, so $$\int y (f^{-1})'(y) \, dy = \int f(f^{-1}(y)) (f^{-1})'(y) \, dy = \int (F \circ f^{-1})'(y) \, dy = (F \circ f^{-1})(y) + C .$$ This formula, in the general case, seems to be surprisingly new: Wikipedia gives 1905 as the discovery date (and a pile of other references on extensions to worse $$f^{-1}$$).

• Whoa! So that's what was going on! – subrosar May 6 '20 at 21:21
• Yes, I was pretty surprised when I found that Wikipedia page (or was it in Spivak?), several years after seeing all the suspiciously-similar formulae for integrating the inverse trigonometrical functions! – Chappers May 6 '20 at 21:23

For example

$$\int \log^2x\,dx=x\log^2x-2\int\log x\,dx=x\log^2x-2x\log x+2x+C$$ or also

$$\int\arcsin x\,dx=x\arcsin x+\sqrt{1-x^2}+C$$

But, in some way, the above are odd-looking variations over the same theme...

The error function, $$\operatorname{erf}$$, is defined by $$\operatorname{erf}(x) = \frac{2}{\sqrt \pi} \int_0^x e^{-t^2} dt.$$ Of course, it is not inmediate to think an antiderivative for this, but for the FTC we easily see that if $$u(x) = \operatorname{erf}(x)$$, then $$u'(x) = \frac{2}{\sqrt \pi} e^{-x^2}$$ and then, using the same trick, $$\int \operatorname{erf}(x) dx = x \operatorname{erf}(x) - \frac{1}{\sqrt \pi} e^{-x^2} +c.$$