Integral of the form $\int_a^b f^{-1}(x)\,dx$ I desire to evaluate $$I=\int_1^{e+1}f^{-1}(t)\,dt$$ where $f(x)=x+\log x\, $ 
My try was to let $t=f(u)\,$ in order to get $$I=\int uf'(u)\,du=uf(u)-\int f(u)\,du$$  and the integral was pretty much over, however I dont know how to get the new bounds after substituting and finding the inverse I think is pretty hard. I would appreciate some guidance.
 A: Hint: 
If $f$ is monotonic
$$\int_{f(a)}^{f(b)} f^{-1}(x) \, dx + \int_a^b f(x) \, dx = bf(b) - af(a),$$
and $f(1) = 1, f(e) = e + 1$
A: For the lower bound, you have that $t=1$. Since $t=f(u)$, you can set those equal, and you get $f(u)=1$. Since $f(x) =x+\log(x)$, you have $u+\log(u)=1$, and by inspection, $u=1$. For the upper bound, it's more complicated if you actually have $\log(x)$, but if you instead have $\ln(x)$, then the upper bound is $e$. (Note that $\log$, with no qualification, is generally understood to mean log base 10; if you want log base $e$, it's best to write as $\ln$ to avoid confusion).
Both the manipulation of the bounds, and solving for $u$, is simplified by the fact that $f$ is monotonic. If you had a non-monotonic function, much of what you did, and much of my answer, would not be valid.
Another way of looking at this is that integral can be interpreted as the area under a curve. Taking the inverse switches $x$ and $y$. So instead of finding the area by going from left to right and taking the height, you can go from bottom to top and take the width.
A: The formula is
$$\int_{f(a)}^{f(b)}f^{-1}\left(x\right)dx=\left[xf\left(x\right)\right]_a^b-\int_a^bf\left(x\right)dx$$
