Question

If $\int \frac{x^2-x+1}{{(x^2+1)}^{\frac{3}{2}}}{e}^{x}dx=e^{x}f\left(x\right)+c,$ then which of the following is/are true:

• A. $f\left(x\right)$ is an even function
• B. $f\left(x\right)$ is an odd function
• C. The range of $f\left(x\right)$ is $(0,1]$
• D. $f\left(1\right)=\frac{1}{2}$

Answer 1

Answer: (A), (C)

The range of $f\left(x\right)$ is $(0,1]$

$I=\int \frac{x^2-x+1}{{(x^2+1)}^{\frac{3}{2}}}{e}^{x}dx=\int e^x[\frac{x^2+1}{{(x^2+1)}^{\frac{3}{2}}}-\frac{x}{{(x^2+1)}^{\frac{3}{2}}}]dx=\int e^x[\frac{1}{\sqrt{x^2+1}}+\left\{\frac{-x}{{(x^2+1)}^{\frac{3}{2}}}\right\}]dx\underset{f\left(x\right)}{\downarrow }\underset{f^{\prime }\left(x\right)}{\downarrow }=\int e^x[f\left(x\right)+f^{\prime }\left(x\right)]dx,\text{where}f\left(x\right)=\frac{1}{\sqrt{x^2+1}}=e^{x}f\left(x\right)+c=\frac{e^x}{\sqrt{x^2+1}}+c$
$\Rightarrow f\left(1\right)=\frac{1}{\sqrt{2}}$
$\because f\left(x\right)=f(-x)=\frac{1}{\sqrt{x^2+1}}$
$\Rightarrow f\left(x\right)$ is even function and the range of $\frac{1}{\sqrt{x^2+1}}$ is $(0,1]$