Question

The primitive of the function
$f\left(x\right)=\left(1-\frac{1}{x^2}\right)a^{x+\frac{1}{x}},a>0\mathrm{is}$

(a) $\frac{a^{x+{\displaystyle \frac{1}{x}}}}{{\log }_{e}a}$

(b) ${\log }_{e}a·a^{x+\frac{1}{x}}$

(c) $\frac{a^{x+{\displaystyle \frac{1}{x}}}}{x}{\log }_{e}a$

(d) $x\frac{a^{x+{\displaystyle \frac{1}{x}}}}{{\log }_{e}a}$

Answer 1

(a) $\frac{a^{x+{\displaystyle \frac{1}{x}}}}{{\log }_{e}a}$

$$\begin{gathered} f\left(x\right)=\left(1-\frac{1}{x^2}\right)·a^{x+\frac{1}{x}} \\ \therefore \int f\left(x\right)dx=\int \left(1-\frac{1}{x^2}\right)·a^{x+\frac{1}{x}}dx \\ \mathrm{Let}\left(x+\frac{1}{x}\right)=t \\ \Rightarrow \left(1-\frac{1}{x^2}\right)dx=dt \\ \therefore \int f\left(x\right)dx=\int a^t·dt \\ =\frac{a^t}{{\log }_{e}a}+C \\ =\frac{a^{x+{\displaystyle \frac{1}{x}}}}{{\log }_{e}a}+C\mathit{}\mathit{}\mathit{}\mathit{}\mathit{}\mathit{}\mathit{}\mathit{}\mathit{}\mathit{}\left(\mathit{\because }t\mathit{}=x+\frac{1}{x}\right) \end{gathered}$$