Integral Calculus Question 246
Question: $ \int_{{}}^{{}}{x\sqrt{2x+3}}\ dx= $
[AISSE 1985]
Options:
A) $ \frac{x}{3}{{(2x+3)}^{3/2}}-\frac{1}{15}{{(2x+3)}^{5/2}}+c $
B) $ \frac{x}{3}{{(2x+3)}^{3/2}}+\frac{1}{15}{{(2x+3)}^{5/2}}+c $
C) $ \frac{x}{2}{{(2x+3)}^{3/2}}+\frac{1}{6}{{(2x+3)}^{5/2}}+c $
D) None of these
Show Answer
Answer:
Correct Answer: A
Solution:
$ \int_{{}}^{{}}{x{{(2x+3)}^{1/2}}dx} $
$ =x\frac{{{(2x+3)}^{3/2}}}{3/2}\frac{1}{2}-\int_{{}}^{{}}{\frac{{{(2x+3)}^{3/2}}}{3/2}\frac{1}{2},dx+c} $
$ =\frac{1}{3}x{{(2x+3)}^{3/2}}-\frac{1}{3}\int_{{}}^{{}}{{{(2x+3)}^{3/2}}dx+c} $
$ =\frac{1}{3}x{{(2x+3)}^{3/2}}-\frac{1}{15}{{(2x+3)}^{5/2}}+c. $
BETA
