How do you find the antiderivative of #int (6x)/(x^2-7)^(1/9)dx# from #[3,4]#?

1 Answer
Noah G
Feb 1, 2017

The integral is approximately equal to #17.546#.

Explanation:

The first thing to do with definite integrals is to make sure that they're in fact definite and not improper.

The integral #(6x)/(x^2 -7)^(1/9)# is continuous on #[3,4]#. We are dealing with a definite integral.

I think we should consider a u-substitution to integrate. Let #u = x^2 - 7#. Then #du = 2xdx# and #dx = (du)/(2x)#. Furthermore, the new bounds of integration become #2# to #9# because we will now be working in #u#.

#=>int_2^9 (6x)/u^(1/9) * (du)/(2x)#

#=>int_2^9 3/u^(1/9)#

#=>3int_2^9 u^(-1/9)#

#=>3[9/8u^(8/9)]_2^9#

#=>3[9/8(9)^(8/9) - 9/8(2)^(8/9)]#

#~~17.546#

Hopefully this helps!

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