If #int 1/(x^22(x^7-6)) dx = A { ln(k)^6 + 9k^2 -2k^3 - 18k} +c# What is the value of 'A' and 'k'?

This is actually a MCQ . The options available are as follows:

(a) A= #1/9072#, k= #((x^7-6)/x^7)#

(b) A= #1/54432#, k= #((x^7-6)/x^7)#

(c) A= #1/54432#, k= #(x^7/(x^7-6))#

(d) none .

I am supposed to solve this in under 3 minutes and in a limited workspace with no calculator.
So, a brief solution or an approach to identify the answer without a lot of steps would be heavily appreciated.
Thank You!

1 Answer
Ananda Dasgupta
Mar 16, 2018

#k = (x^7-6)/x^7# and #A = 1/(42 times 6^4)=1/54432#

Explanation:

#x^22 (x^7-6) =x^29 (x^7-6)/x^7#

Put #k = (x^7-6)/x^7=1-6/x^7#, #implies dk = 42/x^8dx# and #x^7 = 6/(1-k)#.

Thus

#int dx/(x^22 (x^7-6))=int dx/(x^29 (x^7-6)/x^7) = 1/42 int {dk}/((6/(1-k))^3k) #
#= 1/(42times 6^3)int (1/k-3+3k-k^2)dk #
#= 1/(42 times 6^3)(ln k-3k+3/2k^2-k^3/3)#
#= 1/(42 times 6^4)(ln k^6-18k+9k^2-2k^3)#

This is in the form of the answer provided, with

#k = (x^7-6)/x^7# and #A = 1/(42 times 6^4)=1/54432#

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