`int ((x + 6)/(sqrt(x))) dx` Find the indefinite integral.

Textbook Question

Chapter 4, 4.1 - Problem 21 - Calculus of a Single Variable (10th Edition, Ron Larson).
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gsarora17's profile pic

gsarora17 | (Level 2) Associate Educator

Posted on

`int((x+6)/sqrt(x))dx`

apply the sum rule

`=int(x/sqrt(x))dx+int(6/sqrt(x))dx`

Now,

`int(x/sqrt(x))dx=intsqrt(x)dx`

apply the power rule

`=x^(1/2+1)/(1/2+1)`

`=2/3x^(3/2)`

`int(6/sqrt(x))dx=6intx^(-1/2)dx`

`=6(x^(-1/2+1)/(-1/2+1))`

`=6(x^(1/2)/(1/2))`

`=12sqrt(x)`

`:.int((x+6)/sqrt(x))dx=(2x^(3/2))/3+12sqrt(x)+C`

C is constant

scisser's profile pic

scisser | (Level 3) Honors

Posted on

Simplify by dividing each term in the numerator by the denominator

`int(x/sqrtx)dx+int(6/sqrtx)dx`

Use the antiderivative rule: int (a^n=(a^(n+1))/(n+1))

`=(2x^(3/2))/3+12sqrtx+C`

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