Integral Calculus Question 457
Question: $ \int_{{}}^{{}}{(x+3){{(x^{2}+6x+10)}^{9}}\ dx} $ equals
[SCRA 1996]
Options:
A) $ \frac{1}{20}{{(x^{2}+6x+10)}^{10}}+c $
B) $ \frac{1}{20}{{(x+3)}^{2}}{{(x^{2}+6x+10)}^{10}}+c $
C) $ \frac{1}{16}{{(x^{2}+6x+10)}^{8}}+c $
D) $ \frac{1}{38}{{(x+3)}^{19}}+\frac{1}{2}(x+3)+c $
Show Answer
Answer:
Correct Answer: A
Solution:
$ \int_{{}}^{{}}{(x+3){{(x^{2}+6x+10)}^{9}}dx} $ $ =\frac{1}{2}\int_{{}}^{{}}{(2x+6){{(x^{2}+6x+10)}^{2}}dx} $ $ =\frac{1}{2}\frac{{{(x^{2}+6x+10)}^{10}}}{10}+c $ $ =\frac{1}{20}{{(x^{2}+6x+10)}^{10}}+c $ .
BETA
