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.
 A: 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  $$\int f^{-1}(x)dx=f^{-1}(x)x-g(y)=xf^{-1}(x)-g(f^{-1}(x))$$.
A: 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.
A: 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. 
A: 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.
Hope it adds something to this thread!
