How do you find #\int ( \sqrt { x } + \root[ 7] { x } ) d x#?

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Answer 1 · 2017-05-23T07:34:28

#2/3x^(3/2)+7/8x^(8/7)+C#

firstly change the roots to power form

#int(sqrtx+root7x)dx#

#=int(x^(1/2)+x^(1/7))dx#

to integrate this we use the power rule for integration:

#int(x^n)dx=(x^(n+1)/(n+1))+C, n!=-1#

#=int(x^(1/2)+x^(1/7))dx=(x^(3/2)/(3/2))+(x^(8/7)/(8/7))+C#

we now tidy up.

#2/3x^(3/2)+7/8x^(8/7)+C#