Integration by parts leads to a rule for integrating inverses that usually gives good results:
$$\begin{aligned} \int f^{-1}(x) d x &=\int y f^{\prime}(y) d y \quad \quad y=f^{-1}(x), \quad x=f(y) \\ &=y f(y)-\int f(y) d y \quad u=y, d v=f^{\prime}(y) d y \\ &=x f^{-1}(x)-\int f(y) d y \end{aligned}$$
The idea is to take the most complicated part of the integral, in this case $f^{-1}(x),$ and simplify it first. For the integral of $\ln x,$ we get
$$\begin{aligned} \int \ln x d x &=\int y e^{y} d y \\ &=y e^{y}-e^{y}+C \\ &=x \ln x-x+C \end{aligned} \quad \begin{aligned} y &=\ln x, \quad x=e^{y} \\ d x &=e^{y} d y \end{aligned}$$
For the integral of $\cos ^{-1} x$ we get
$$\begin{aligned} \int \cos ^{-1} x d x &=x \cos ^{-1} x-\int \cos y d y \\ &=x \cos ^{-1} x-\sin y+C \\ &=x \cos ^{-1} x-\sin \left(\cos ^{-1} x\right)+C \end{aligned}$$
Use the formula
$$\int f^{-1}(x) d x=x f^{-1}(x)-\int f(y) d y \quad y=f^{-1}(x)$$
to evaluate the integrals in Exercises $67-70 .$ Express your answers in terms of $x .$
$$\int \log _{2} x d x$$
Another way to integrate $f^{-1}(x)$ (when $f^{-1}$ is integrable, of course) is to use integration by parts with $u=f^{-1}(x)$ and $d v=d x$ to rewrite the integral of $f^{-1}$ as
$$\int f^{-1}(x) d x=x f^{-1}(x)-\int x\left(\frac{d}{d x} f^{-1}(x)\right) d x$$