Question

$\int 5^{mx}{7}^{nx}dx,m,n\in N$ is equal to

• A. $\frac{5^{mx}+7^{nx}}{m\log 5+n\log 7}+K$
• B. $\frac{e^{(m\log 5+n\log 7)x}}{\log 5^m+\log 7^n}+K$
• C. $\frac{(m.n)5^{mx}+7^{nx}}{m\log 5+n\log 7}+K$
• D. None of these

Answer 1

Answer: (A), (B)

The correct options are
A $\frac{5^{mx}+7^{nx}}{m\log 5+n\log 7}+K$
B $\frac{e^{(m\log 5+n\log 7)x}}{\log 5^m+\log 7^n}+K$

$\int 5^{mx}{7}^{nx}dx=\int (e^{\log 5}{)}^{mx}\cdot (e^{\log 7}{)}^{nx}=\int e^{x(m\log 5+n\log 7)}dx(\because a^x=e^{{\log }_{a}x}=e^{x\log a})$

$=\frac{e^{x(m\log 5+n\log 7)}}{(m\log 5+n\log 7)}+K=\frac{5^{mx}+7^{nx}}{\log (5^m.7^n)}+K$ since $\int e^{ax}=\frac{e^{ax}}{a}$

On simplifying we have:

$=\frac{5^{mx}+7^{nx}}{mlog5+nlog7}+K$

Thus option A is correct.