Question

If $\int \frac{(7{)}^{x^{\frac{1}{2}}}}{x^3}dx=k(7{)}^{x^{\frac{1}{2}}}+c$ then k$=$ ___________.

• A. $\frac{-1}{2lo{g}_{e}7}$
• B. $\frac{-1}{7lo{g}_{e}7}$
• C. $\frac{1}{7}$
• D. $\frac{1}{lo{g}_{e}7}$

Answer 1

Answer: (A)

$\frac{-1}{2lo{g}_{e}7}$

$I=\int$ $\frac{7^{\frac{1}{x^2}}}{x^3}dx$
$=\int \left(7^{\frac{1}{x^2}}\right)\left(\frac{1}{x^3}\right)dx$
Now Taking $\frac{1}{x^2}=t$
$\therefore$ We get $\frac{1}{x^3}dx=-\frac{1}{2}dt$
$I=-\frac{1}{2}\int 7^{t}dt$
$=-\frac{1}{2}\left(7^t\right)log7+c$
$I=\frac{-1}{2log7}\left(7^{\frac{1}{x^2}}\right)+c$
Now Compare with the given result
$\therefore k=\frac{-1}{2log7}$.