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

Question

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!

Answer 1

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.

Answer 2

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...

Answer 3

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 $f^{-1}$).

Answer 4

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, $$\int \operatorname{erf}(x) dx = x \operatorname{erf}(x) - \frac{1}{\sqrt \pi} e^{-x^2} +c.$$