# Integral dx/sqrt(x - a)integral dx/sqrt(1/ax)

#### rebeka

integral dx/sqrt(x - a)
integral dx/sqrt(1/ax)

Last edited:

#### rebeka

sorry for wasting time I think I just understood substitution, out of embarassment im never coming back to this forum(bright side of life I learned more in three days than I have in two years)

#### HallsofIvy

Homework Helper
We all embarrass ourselves at one time or another! (Some of us more than others. Believe me, I know about embarrassing myself!) PLEASE come back to PhysicsForums!

(I need someone to make me look good!!!)

Anyway, for those who are still wondering how to do these:

$$\int \sqrt{x-a}dx$$

Let u= x-1 so du= dx and $$\sqrt{x-a}= \sqrt{u}= u^{\frac{1}{2}}$$. The integral becomes $$\int u^{\frac{1}{2}}du$$ which can be done with the power law.

$$\int \frac{dx}{\sqrt{\frac{1}{ax}}}$$

$$\frac{1}{\sqrt{\frac{1}{ax}}}$$ is just $$\sqrt{ax}$$.

Let u= ax so du= adx, dx= (1/a)du and $$\sqrt{ax}= \sqrt{u}= u^{\frac{1}{2}}$$

The integral becomes $$\frac{1}{a}\int u^{\frac{1}{2}}du$$

Last edited by a moderator:

#### Orion1

HallsofIvy said:
$$\int \sqrt{x-a}dx$$

HallsofIvy, should not this equation be:
$$\int \frac{dx}{\sqrt{x - a}} = \int u^{-\frac{1}{2}} du$$

#### HallsofIvy

Homework Helper
Oh, you're right! I missed the "1/" in the first post. Of course, the substitution would be exactly the same and the u-integral what you show. Thanks.

### Physics Forums Values

We Value Quality
• Topics based on mainstream science
• Proper English grammar and spelling
We Value Civility
• Positive and compassionate attitudes
• Patience while debating
We Value Productivity
• Disciplined to remain on-topic
• Recognition of own weaknesses
• Solo and co-op problem solving