How do you integrate by substitution #int (x^2+3x+7)/sqrtxdx#?

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Answer 1 · 2017-03-19T03:53:46

#(2/5)x^(5/2) +2x^(3/2) +14sqrtx +C#

Let u = x

#int (u^2+3u +7)/sqrt(u)#

Make it easier by simplifying integrals

#int u^2/sqrt(u)# + #int (3u)/sqrtu# + #int 7/sqrt(u)#

Simplify

#int u^(3/2)# + #int 3sqrtu# + #int 7u^(-1/2)#

Integrate

#2/5*u^(5/2) + 3*(2/3)u^(3/2) + 7*2sqrtu#

Sub back in x

#(2/5)x^(5/2) +2x^(3/2) +14sqrtx +C#