integral involving invertible functions

Solution 1:

$$\int_{a}^{b}(f(x)+xf'(x))dx$$

Let $y=xf(x)$

$$\frac{dy}{dx}=xf'(x)+f(x)$$

Thus our integral becomes,

$$\int_{x=a}^{x=b}dy.$$

$$bf(b)-af(a)$$

The increasing property seems misleading, there is no need for it.

As per OP's request here is a graphical explanation from here by David Mitra.

This is so brilliant, no explanation required.

Solution 2:

Hint: Since $y=f(x)$ is invertible, the function $x=f^{-1}(y)$ is defined, $x=a\iff y=f(a)$ and $x=b\iff y=f(b)$. Thus $$\int_{f(a)}^{f(b)} f^{-1}(y)dy=\int_{f(a)}^{f(b)} xdy=\int_a^b (x \frac{dy}{dx}) dx=\int_a^b x f'(x) dx$$ Now, it is clear that $(xf(x))'=f(x)+xf'(x)$ and hence it is easy to complete.