Integral Calculus Question 225
Question: If $ \int{\frac{dx}{x^{22}(x^{7}-6)}} $ $ =A{In{{(p)}^{6}}+9p^{2}-2p^{3}-18p}+c $ then
Options:
A) $ A=\frac{1}{9072},p=( \frac{x^{7}-6}{x^{7}} ) $
B) $ A=\frac{1}{54432},p=( \frac{x^{7}-6}{x^{7}} ) $
C) $ A=\frac{1}{54432},p=( \frac{x^{7}}{x^{7}-6} ) $
D) $ A=\frac{1}{9072},p={{( \frac{x^{7}-6}{x^{7}} )}^{-1}} $
Show Answer
Answer:
Correct Answer: B
Solution:
[b] Let $ I=\int{\frac{dx}{x^{29}( 1-\frac{6}{x^{7}} )}} $ Put $ 1-\frac{6}{x^{7}}=p\Rightarrow \frac{42}{x^{8}}dx=dp $ and $ x^{7}=\frac{6}{1-p} $
$ \therefore I=\frac{1}{42}\int{\frac{{{(1-p)}^{3}}}{{{(6)}^{3}}p}dp} $ $ =\frac{1}{(42)(216)}\int{\frac{1-p^{3}-3p+3p^{2}}{p}dp} $ $ =\frac{1}{9072}\int{( \frac{1}{p}-p^{2}-3+3p )dp} $ $ =\frac{1}{9072}( \log p-\frac{p^{3}}{3}-3p+\frac{3}{2}p^{2} )+c $ $ =\frac{1}{54432}(6l,np-2p^{3}-18p+9p^{2})+c $ $ =\frac{1}{54432}(l,np^{6}+9p^{2}-2p^{3}-18p)+c $ $ A=\frac{1}{54432},p=( \frac{x^{7}-6}{x^{7}} ) $
BETA
