# If $\int f(x)dx =g(x)$ then $\int f^{-1}(x)dx$ is equal to

If $$\int f(x)dx =g(x)$$ then $$\int f^{-1}(x)dx$$ is equal to

(1) $$g^{-1}(x)$$

(2) $$xf^{-1}(x)-g(f^{-1}(x))$$

(3) $$xf^{-1}(x)-g^{-1}(x)$$

(4) $$f^{-1}(x)$$

My approach is as follows: Let $$f(x)=y$$, therefore $$f^{-1}(y)=x$$, $$\int f^{-1}(f(x))dx =g(f(x))$$

On differentiating we get $$x=g'(f(x))f'(x)$$

After this step, I am not able to proceed.

• Why not try an example to see whether any of those formulas work? $\int x^2\,dx=x^3/3+C,$\int\sqrt x\,dx=(2/3)x^{3/2}+C\$, do any of the formulas work in this case? Jul 11, 2020 at 6:51

There's a nice visual computation of the antiderivative of an inverse function: $$F(x) := \int_0^x f^{-1}(t) dt$$ is an antiderivative for $$f^{-1}(x)$$, and for $$x = a$$, $$F(a)$$ is equal to the green area in the picture below:

If we could figure out the blue area, we would be set, because

\begin{align*} \text{ (green area) } &= \text{ (rectangle area) } - \text{ (blue area) } \\ F(a) &= af^{-1}(a) - \text{ (blue area) }.\ \end{align*}

But if we reflect this picture across the line $$y = x$$, we see the blue area is just the antiderivative of $$f$$, namely $$g(x) := \int_0^x f(t) dt$$, evaluated at $$f^{-1}(a)$$:

So we get $$\text{ (blue area) } = g(f^{-1}(a)),$$

and plugging this in, we get that $$F(a)$$ is equal to the second of the four choices:

\begin{align*} F(a) &= af^{-1}(a) - \text{ (blue area) } \\ &= af^{-1}(a) - g(f^{-1}(a)). \ \end{align*}

† To make the picture look nice, we assumed $$f(0) = f^{-1}(0) = 0$$ without loss of generality.

• Very nice explaination Jul 12, 2020 at 3:09

Ignoring the constant of integration the answer is (2):$$\int f^{-1}(x)dx=\int yf'(y)dy=yf(y)-\int f(y)dy$$ (where I have used integration by parts). Hence $$\int f^{-1}(x)dx=f^{-1}(x)x-g(y)=xf^{-1}(x)-g(f^{-1}(x))$$.

Given $$y=f(x) \implies x=f^{-1}(y)$$ Then $$\int f^{-1}(x) dx= \int f^{-1}(y) dy= \int x dy= \int x \frac{dy}{dx} dx=\int xf'(x) dx$$ $$=xf(x)-\int f(x)dx=xf(x)-g(x)+C.$$ Lastly, we have done integration by parts.

• It’s a very clear answer. You have made every step so clear. Jul 12, 2020 at 3:46

All the other answers have already given you the finest methods of solving the problem. I want to say that as you’re given the options for the integral $$\int f^{-1} (x) dx$$, it would be a nice thing to differentiate the options to see if we get $$f^{-1}(x)$$ (it should follow from The Fundamental Theorem of Calculus ).

Let’s try the first option: $$If ~~~~ \int f^{-1}(x) dx = g^{-1} (x) \\ then~~~~~ f^{-1} (x)= \frac{d}{dx} g^{-1} (x) \\ f^{-1} (x)= \left( \frac{d~g(x)}{dx} \right)^{-1}\\ f^{-1} (x)= \frac{1}{f(x)}$$ (in the third step I once again used the FTC for the function$$f(x)$$) the last equality is not true in general, therefore this option is not valid.

Let’s try the second option: $$If~~~~ \int f^{-1}(x) dx = xf^{-1} (x) - g\left( f^{-1}(x)\right) \\ then~~~~ f^{-1}(x)= \frac{d}{dx} \left[xf^{-1} (x) - g\left( f^{-1}(x)\right) \right] \\ f^{-1} (x)= f^{-1}(x) + x \frac{d~f^{-1}(x)}{dx} - f\left(f^{-1} (x)\right) \frac{d~f^{-1}(x)}{dx}\\ f^{-1}(x) = f^{-1}(x)$$ Hence, the second option is correct.