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

Question

#int (x^2+3x+7)/sqrtxdx#

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Answer 1 · 2018-01-23T10:41:40

#2/5sqrt(x)*(x^(4)+5x+35)+C#

We have the integral:

#int(x^2+3x+7)/sqrt(x)dx#

Rewrite in terms of the rule of addition of fractions

#int(x^2)/sqrt(x)+(3x)/sqrt(x)+7/sqrt(x)dx#

Use the quotient rule for exponents #a^n/a^m=a^(n-m)#

#intx^(3/2)+3x^(1/2)+7x^(-1/2)dx#

Use the sum rule for integrals #int(f+g)dx=intfdx+intgdx#

#intx^(3/2)dx+int3x^(1/2)dx+int7x^(-1/2)dx#

Integrate by the power rule #intx^ndx=1/(n+1)*x^(n+1)#

#2/5x^(5/2)+3*2/3x^(3/2)+2*7x^(1/2)+C#

Simplify

#2/5x^(5/2)+2x^(3/2)+14x^(1/2)+C#

Or

#2/5sqrt(x)*(x^(4)+5x+35)+C#