1. Consider the function $f:(-1, \infty) \rightarrow \mathbb{R}, f(x)=\frac{\sqrt{x^{2}+1}}{x+1}$.

a) Calculate $f^{\prime}(x), x>-1$.

b) Calculate $\int_{0}^{1} \frac{x^{2}+3 x}{(x+1)^{2} \sqrt{x^{2}+1}} d x$.
## Solution.

a) For $x>-1$, we have $f^{\prime}(x)=\frac{x-1}{(x+1)^{2} \sqrt{x^{2}+1}}$. (2p)

b) We have

$$
\begin{aligned}
\int_{0}^{1} \frac{x^{2}+3 x}{(x+1)^{2} \sqrt{x^{2}+1}} d x & =\int_{0}^{1} \frac{(x+1)^{2}+(x-1)}{(x+1)^{2} \sqrt{x^{2}+1}} d x(\mathbf{p}) \\
& =\int_{0}^{1} \frac{1}{\sqrt{x^{2}+1}} d x+\int_{0}^{1}\left(\frac{\sqrt{x^{2}+1}}{x+1}\right)^{\prime} d x(\mathbf{p} \mathbf{)} \\
& =\left.\ln \left(x+\sqrt{x^{2}+1}\right)\right|_{0} ^{1}+\left.\frac{\sqrt{x^{2}+1}}{x+1}\right|_{0} ^{1}(\mathbf{1} \mathbf{p}) \\
& =\ln (1+\sqrt{2})+\frac{\sqrt{2}}{2}-1 .(\mathbf{1} \mathbf{p})
\end{aligned}
$$