Integral Calculus Question 135
Question: $ \int_{{}}^{{}}{{a^{3x+3}}dx}= $
[Roorkee 1977]
Options:
A) $ \frac{{a^{3x+3}}}{\log a}+c $
B) $ \frac{{a^{3x+3}}}{3\log a}+c $
C) $ {a^{3x+3}}\log a+c $
D) $ 3{a^{3x+3}}\log a+c $
Show Answer
Answer:
Correct Answer: B
Solution:
Put $ t=3x+3\Rightarrow dt=3,dx, $ then $ \int_{{}}^{{}}{{a^{3x+3}}dx}=\frac{1}{3}\int_{{}}^{{}}{a^{t}dt}=\frac{1}{3}\frac{a^{t}}{{\log_{e}}a}+c=\frac{{a^{3x+3}}}{3{\log_{e}}a}+c $ .
BETA
