Question

If $\int \frac{x^2-x+1}{(x^2+1{)}^{3/2}}{e}^{x}dx=e^{x}f\left(x\right)+c$, then

• A. $f\left(x\right)$ is an even function
• B. $f\left(x\right)$ is a bounded function
• C. the range of $f\left(x\right)$ is $(0,1]$
• D. $f\left(x\right)$ has two points of extrema

Answer 1

Answer: (A), (B), (C)

The correct options are
A $f\left(x\right)$ is an even function
B $f\left(x\right)$ is a bounded function
C the range of $f\left(x\right)$ is $(0,1]$

$I=\int \frac{x^2-x+1}{(x^2+1{)}^{3/2}}{e}^{x}dx$
$=\int e^x[\frac{x^2+1}{(x^2+1{)}^{3/2}}-\frac{x}{(x^2+1{)}^{3/2}}]dx$
$=\int e^x[\frac{1}{\sqrt{x^2+1}}+\left\{\frac{-x}{(x^2+1{)}^{3/2}}\right\}]dx$
$\int e^x\left[f\right(x)+f^{\prime }(x\left)\right]dx$, where $f\left(x\right)=\frac{1}{\sqrt{x^2+1}}$
$=e^{x}f\left(x\right)+c=\frac{e^x}{\sqrt{x^2+1}}+c$

The graph of $f\left(x\right)$ is given in the figure.
From the graph, $f\left(x\right)$ is even, bounded function and has the range $(0,1]$.