integral of inverse functions

Let \[dy=f^{\prime}(x)\,dx\]. Then substitute \[y=f(x)\] and \[dy=f^{\prime}(x)\,dx\] to obtain \[\int f^{-1}(y)\,dy=\int f^{-1}(f(x))f^{\prime}(x)\,dx=\int xf^{\prime}(x)\,dx\].

Now apply integration by parts, let \[u(x)=x\] and \[v^{\prime}(x)=f^{\prime}(x)\]. Then \[u^{\prime}(x)=1\] and \[v(x)=\int f^{\prime}(x)\,dx=f(x)+C\]. Using the formula,

\begin{align*} \int xf^{\prime}(x)\,dx&=x\left( f(x)+C \right)-\int 1\cdot \left( f(x)+C \right)\,dx\\ &=xf(x)+Cx-\left( F(x)+Cx \right)\\ &=xf(x)-F(x)\\ &=xf(x)-\int f(x)\,dx\\ \end{align*}

Then with \[xf(x)-\int f(x)\,dx\] we substitute \[y\] into it to get \[xf(x)-\int f(x)\,dx=yf^{-1}(y)-F(f^{-1}(y))\].

However, since \[C\] is an arbitrary constant we usually ignore it while writing the formula.

Let \[y=f(x)=\tan(x)\] and \[x=\tan^{-1}(y)\], then \[\int f(x)\,dx=-\ln \left| \cos(x) \right|\]. Based on the trigonometric identities, we also know that \[\cos(x)=\frac{1}{\sqrt{1+\tan^{2}x}}=\frac{1}{\sqrt{1+y^{2}}}\].

\begin{align*} \int \tan^{-1}(y)\,dy&=y\tan^{-1}(y)-\left( -\ln \left| \cos(\tan^{-1}(y)) \right| \right)\\ &=y\tan^{-1}(y)+\ln \left| \cos(x) \right|\\ &=y\tan^{-1}(y)+\ln \left| \frac{1}{\sqrt{1+y^{2}}} \right|\\ &=y\tan^{-1}(y)+\ln \left| 1+y^{2} \right|^{-\frac{1}{2}}\\ &=y\tan^{-1}(y)-\frac{1}{2}\ln \left| 1+y^{2} \right|\\ \end{align*}

Therefore, \[\int \tan^{-1}(y)\,dy=y\tan^{-1}(y)-\frac{1}{2}\ln \left| 1+y^{2} \right|+C\].