# Thread: Getting Rid of a Constant of 1 in Integral Substitution

1. ## Getting Rid of a Constant of 1 in Integral Substitution

$\displaystyle \int \dfrac{dx}{2\sqrt{x} + 2x}$. - The key here is pulling the constant out, before doing u substitution.

$\displaystyle \int (\dfrac{1}{2})\dfrac{dx}{\sqrt{x} + x}$.

$\displaystyle \int (\dfrac{1}{2})\dfrac{du}{u}$.

$\displaystyle u = \sqrt{x} + x$.

$\displaystyle du = \dfrac{1}{2\sqrt{x}} + 1dx$. ?? What is the next step? Somehow we need to get rid of that constant of 1, and $\displaystyle \dfrac{1}{x}$

Note: The answer to this problem is $\displaystyle \ln(\sqrt{x} + x) + C$ But do we get there?

2. ## Re: Getting Rid of a Constant of 1 in Integral Substitution

Originally Posted by Jason76
$\displaystyle \int \dfrac{dx}{2\sqrt{x} + 2x}$. - The key here is pulling the constant out, before doing u substitution.

$\displaystyle \int (\dfrac{1}{2})\dfrac{dx}{\sqrt{x} + x}$.

$\displaystyle \int (\dfrac{1}{2})\dfrac{du}{u}$.

$\displaystyle u = \sqrt{x} + x$.

$\displaystyle du = \dfrac{1}{2\sqrt{x}} + 1dx$. ?? What is the next step? Somehow we need to get rid of that constant of 1, and $\displaystyle \dfrac{1}{x}$

Note: The answer to this problem is $\displaystyle \ln(\sqrt{x} + x) + C$ But do we get there?
That technique isn't going to work. Try the substitution $\displaystyle y = \sqrt{x}$ and see where that gets you.

-Dan

3. ## Re: Getting Rid of a Constant of 1 in Integral Substitution

Unfortunately, (it breaks my heart to tell you this), but the answer (you have) is wrong!

$\displaystyle \int_a^b \frac{1}{2\sqrt(x) + 2x } dx = ln|\sqrt(x) +1|$

I'll give you a critical step.

Factor out in the denominator $\displaystyle \sqrt(x)$

Then use the substitution $\displaystyle u = \sqrt(x)$

See where that gets you =)