Evaluate the integral of (x√x-3dx)?

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Answer 1 · 2018-05-09T17:24:24

the answer
#intx^(3/2)-3*dx=[2/5*x^(5/2)-3x]+c#

if your integral was #intxsqrtx-3dx#

rewrite it #intx^(3/2)-3*dx#

now we will evaluate it

#intx^(3/2)-3*dx=[2/5*x^(5/2)-3x]+c#

Answer 2 · 2018-05-09T18:17:10

# 2/5(x-3)^(3/2)*(x+2)+C#.

If the integral is, #I=intxsqrt(x-3)dx#, then, we have,

#I=int{(x-3)+3}sqrt(x-3)dx#,

#=int{(x-3)sqrt(x-3)+3sqrt(x-3)}dx#,

#=int{(x-3)^(3/2)+3(x-3)^(1/2)}dx#,

#=(x-3)^(3/2+1)/(3/2+1)+3*(x-3)^(1/2+1)/(1/2+1)#,

#=2/5(x-3)^(5/2)+3*2/3(x-3)^(3/2)#,

#=2/5(x-3)^(3/2){(x-3)+5}#.

#:. I=intxsqrt(x-3)dx=2/5(x-3)^(3/2)*(x+2)+C#.

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