Simplifying integral: How to simplify 2sqrt{bx+a)[(bx+a)-3a] to 2sqrt{bx+a}[bx-2a] ?

[flyingfishcattle]

Hi,

I have evaluated the integral for:

. . . . .\(\displaystyle \displaystyle \int\, \dfrac{x}{\sqrt{a\, +\, bx\,}}\, dx\)

For my answer I got:

. . . . .\(\displaystyle \displaystyle \dfrac{2\, (bx\, +\, a)^{3/2}}{3b^2}\, -\, \dfrac{2a\, \sqrt{bx\, +\, a\,}}{b^2}\, +\, C\)

But I don't know how to simplify that above to get the below:

. . . . .\(\displaystyle \displaystyle \dfrac{2\, (bx\, -\, 2a)\, \sqrt{bx\, +\, a\,}}{3b^2}\, +\, C\)

Here is my work so far:

. . . . .\(\displaystyle \begin{align}\displaystyle \dfrac{2\, (bx\, +\, a)^{3/2}}{3b^2}\, -\, \dfrac{2a\, \sqrt{bx\, +\, a\,}}{b^2}\, +\, C\, &=\, \dfrac{2\, (bx\, +\, a)^{3/2}}{3b^2}\, -\, \dfrac{3\, \cdot\, 2a\, \sqrt{bx\, +\, a\,}}{3\, \cdot\, b^2}\, +\, C\\
\\
&=\, \dfrac{2\, (bx\, +\, a)^{3/2}\, -\, 3\, \cdot\, 2a\, \sqrt{bx\, +\, a\,}}{3b^2}\\
\\
&=\, \dfrac{\left(2\, \sqrt{bx\, +\, a\,}\right)\, \big((bx\, +\, a)\, -\, 3a\big)}{3b^2}\end{align}\)

How do I get from this to the final form, below?

. . . . .\(\displaystyle \displaystyle \dfrac{2\, (bx\, -\, 2a)\, \sqrt{bx\, +\, a\,}}{3b^2}\, +\, C\)

Any help would be appreciated

 

[Steven G]

Hi,

I have evaluated the integral for:

. . . . .\(\displaystyle \displaystyle \int\, \dfrac{x}{\sqrt{a\, +\, bx\,}}\, dx\)

For my answer I got:

. . . . .\(\displaystyle \displaystyle \dfrac{2\, (bx\, +\, a)^{3/2}}{3b^2}\, -\, \dfrac{2a\, \sqrt{bx\, +\, a\,}}{b^2}\, +\, C\)

But I don't know how to simplify that above to get the below:

. . . . .\(\displaystyle \displaystyle \dfrac{2\, (bx\, -\, 2a)\, \sqrt{bx\, +\, a\,}}{3b^2}\, +\, C\)

Here is my work so far:

. . . . .\(\displaystyle \begin{align}\displaystyle \dfrac{2\, (bx\, +\, a)^{3/2}}{3b^2}\, -\, \dfrac{2a\, \sqrt{bx\, +\, a\,}}{b^2}\, +\, C\, &=\, \dfrac{2\, (bx\, +\, a)^{3/2}}{3b^2}\, -\, \dfrac{3\, \cdot\, 2a\, \sqrt{bx\, +\, a\,}}{3\, \cdot\, b^2}\, +\, C\\
\\
&=\, \dfrac{2\, (bx\, +\, a)^{3/2}\, -\, 3\, \cdot\, 2a\, \sqrt{bx\, +\, a\,}}{3b^2}\\
\\
&=\, \dfrac{\left(2\, \sqrt{bx\, +\, a\,}\right)\, \big((bx\, +\, a)\, -\, 3a\big)}{3b^2}\end{align}\)

How do I get from this to the final form, below?

. . . . .\(\displaystyle \displaystyle \dfrac{2\, (bx\, -\, 2a)\, \sqrt{bx\, +\, a\,}}{3b^2}\, +\, C\)

Any help would be appreciated

First get common denominators and then factor out what is common.
In case you need help with that then show us your work.
What do you think the common denominator would be?

 

[Denis]

You were doing ok up to your last 2 steps: go look carefully.

 

[Dr.Peterson]

Hi,

I have evaluated the integral for:

. . . . .\(\displaystyle \displaystyle \int\, \dfrac{x}{\sqrt{a\, +\, bx\,}}\, dx\)

For my answer I got:

. . . . .\(\displaystyle \displaystyle \dfrac{2\, (bx\, +\, a)^{3/2}}{3b^2}\, -\, \dfrac{2a\, \sqrt{bx\, +\, a\,}}{b^2}\, +\, C\)

But I don't know how to simplify that above to get the below:

. . . . .\(\displaystyle \displaystyle \dfrac{2\, (bx\, -\, 2a)\, \sqrt{bx\, +\, a\,}}{3b^2}\, +\, C\)

Here is my work so far:

. . . . .\(\displaystyle \begin{align}\displaystyle \dfrac{2\, (bx\, +\, a)^{3/2}}{3b^2}\, -\, \dfrac{2a\, \sqrt{bx\, +\, a\,}}{b^2}\, +\, C\, &=\, \dfrac{2\, (bx\, +\, a)^{3/2}}{3b^2}\, -\, \dfrac{3\, \cdot\, 2a\, \sqrt{bx\, +\, a\,}}{3\, \cdot\, b^2}\, +\, C\\
\\
&=\, \dfrac{2\, (bx\, +\, a)^{3/2}\, -\, 3\, \cdot\, 2a\, \sqrt{bx\, +\, a\,}}{3b^2}\\
\\
&=\, \dfrac{\left(2\, \sqrt{bx\, +\, a\,}\right)\, \big((bx\, +\, a)\, -\, 3a\big)}{3b^2}\end{align}\)

How do I get from this to the final form, below?

. . . . .\(\displaystyle \displaystyle \dfrac{2\, (bx\, -\, 2a)\, \sqrt{bx\, +\, a\,}}{3b^2}\, +\, C\)

Any help would be appreciated

But there is no extra work needed between the next to last and last lines! Just combine like terms: bx + a - 3a = bx - 2a. Everything you've done is correct, including that last step.