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How do you integrate #int sqrt(1+2x)# using substitution?

How do you integrate #int sqrt(1+2x)dx# using substitution?

Answer 1

Ezekiel Alexander

#1/3(sqrt(1+2x))^3+C, or#

#1/3(1+2x)^(3/2)+C#.

We subst. #1+2x=t^2" in "I=intsqrt(1+2x)dx#.

#:. 2dx=2tdt, or, dx=tdt#. Also, #sqrt(1+2x)=t#

Therefore, #I=intt*tdt=intt^2dt=t^3/3#

#=1/3(sqrt(1+2x))^3+C, or#

#intI=1/3(1+2x)^(3/2)+C#.

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Answer 2

Ezra Adams

To integrate (\int \sqrt{1 + 2x} , dx) using substitution:

Let (u = 1 + 2x), then (du = 2 , dx).
Solve for (dx) to substitute back into the integral: (dx = \frac{du}{2}).
Substitute (u) and (dx) into the integral: [\int \sqrt{u} \cdot \frac{du}{2}]
Simplify and pull out constants: [\frac{1}{2} \int \sqrt{u} , du]
Integrate (\sqrt{u}): [\frac{1}{2} \cdot \frac{2}{3} u^{\frac{3}{2}} + C]
Substitute back (u = 1 + 2x): [\frac{1}{3} (1 + 2x)^{\frac{3}{2}} + C]

So, (\int \sqrt{1 + 2x} , dx = \frac{1}{3} (1 + 2x)^{\frac{3}{2}} + C), where (C) is the constant of integration.

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Answer from HIX Tutor

When evaluating a one-sided limit, you need to be careful when a quantity is approaching zero since its sign is different depending on which way it is approaching zero from. Let us look at some examples.