SUBIECTUL III Let $I$, $J$ be intervals and $f: I \rightarrow \mathbf{R}^{*}$ a function with a primitive $F: I \rightarrow \mathbf{R}$ that is bijective.

a) Calculate $\int \frac{F^{-1}(x)}{f\left(F^{-1}(x)\right)} d x, x \in J$.

b) Calculate $\int \frac{\ln \left(x+\sqrt{x^{2}+1}\right)}{\sqrt{x^{2}+1}} d x, x \in \mathbf{R}$.
## Solution:

a) From $F$ bijective $\Rightarrow F\left(F^{-1}(x)\right)=x, \forall x \in J$.

By differentiation we obtain

$F^{\prime}\left(F^{-1}(x)\right) \cdot\left(F^{-1}\right)^{\prime}(x)=1, \forall x \in J \Leftrightarrow\left(F^{-1}\right)^{\prime}(x)=\frac{1}{f\left(F^{-1}(x)\right)}, x \in J$

and thus

$\left(\frac{1}{2}\left(F^{-1}(x)\right)^{2}\right)^{\prime}=F^{-1}(x) \cdot\left(F^{-1}(x)\right)^{\prime}=F^{-1}(x) \frac{1}{f\left(F^{-1}(x)\right)}, x \in J$,

therefore: $\int \frac{F^{-1}(x)}{f\left(F^{-1}(x)\right)} d x=\frac{1}{2}\left(F^{-1}(x)\right)^{2}+C, x \in J$.

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## b) Method I:

$$
\int \frac{\ln \left(x+\sqrt{x^{2}+1}\right)}{\sqrt{x^{2}+1}} d x=\int\left(\ln \left(x+\sqrt{x^{2}+1}\right)\right)^{\prime} \cdot \ln \left(x+\sqrt{x^{2}+1}\right) d x=\frac{1}{2} \ln ^{2}\left(x+\sqrt{x^{2}+1}\right)+C
$$

Method II: Let $f(x)=\frac{e^{x}+e^{-x}}{2}, x \in \mathbf{R}$ and $F: \mathbf{R} \rightarrow \mathbf{R}, F(x)=\frac{e^{x}-e^{-x}}{2}$. We have $F^{\prime}=f, F$ bijective and since $F^{-1}(x)=\ln \left(x+\sqrt{x^{2}+1}\right), x \in \mathbf{R} \Rightarrow f\left(F^{-1}(x)\right)=\sqrt{x^{2}+1}$, thus $\int \frac{\ln \left(x+\sqrt{x^{2}+1}\right)}{\sqrt{x^{2}+1}} d x=\int \frac{F^{-1}(x)}{f\left(F^{-1}(x)\right)} d x=\frac{1}{2} \ln ^{2}\left(x+\sqrt{x^{2}+1}\right)+C$.