Integration Problem

[Jason76]

\(\displaystyle \int \dfrac{5x - 2}{\sqrt{x^{2} + 2x}} dx \)

Wanting to convert to the form:

\(\displaystyle u^{n}du = \dfrac{u^{n + 1}}{n + 1} + C \) and \(\displaystyle \int \dfrac{du}{\sqrt u^{2} - a^{2}} = \ln| u + \sqrt{u^{2} - a^{2}}| + C \)

Here is what the book says:

Arrange numerator in the following manner:

\(\displaystyle \int \dfrac{5x - 2}{\sqrt{x^{2} + 2x}} dx = \dfrac{5}{2} \int \dfrac{2x - 4/5 dx}{\sqrt{x^{2} + 2x}}\) What's going on here?

\(\displaystyle \dfrac{5}{2} \int \dfrac{(2x + 2)2x } {\sqrt{x^{2} + 2x}} dx - (\dfrac{5}{2})* (\dfrac{4}{15}) \int \dfrac{dx} {\sqrt{x^{2} + 2x}}\)

I understand the next part of the problem (based on given information from the book):

\(\displaystyle \dfrac{5}{2}\int (x^{2} + 2x)^{-1/2} (2x + 2) dx - 7 \int \dfrac{dx}{\sqrt{(x + 1)^{2} - 1^{2}}}\)

\(\displaystyle 5\sqrt{x^{2} + 2x} - 7\ln|x + 1 + \sqrt{x^{2} + 2x}| + C\) - Final Answer

 

[DrPhil]

\(\displaystyle \int \dfrac{5x - 2}{\sqrt{x^{2} + 2x}} \)

Wanting to convert to the form:

\(\displaystyle u^{n}du = \dfrac{u^{n + 1}}{n + 1} + C \) and \(\displaystyle \int \dfrac{du}{\sqrt u^{2} - a^{2}} = \ln| u + \sqrt u^{2} - a^{2}| + C \)

Show your work!

A possible clue:
\(\displaystyle x^2 + 2x = x^2 + 2x + 1 - 1 = (x + 1)^2 - 1 = u^2 - a^2\)

 

[Jason76]

Show your work!

A possible clue:
\(\displaystyle x^2 + 2x = x^2 + 2x + 1 - 1 = (x + 1)^2 - 1 = u^2 - a^2\)

I understand that part (completing the square). It comes next in the problem. But I don't understand all the "manipulation of constants" that was going on, and "rearranging the numerator" in the beginning parts of the problem.

Shouldn't \(\displaystyle 5x - 2\), at the start of the problem, be divided into two numbers with \(\displaystyle \sqrt{x^{2} + 2x}\) under each number?

 

[soroban]

Hello, Jason76!

\(\displaystyle \displaystyle\int \dfrac{5x - 2}{\sqrt{x^2 + 2x}}\,dx \)

Here is what the book says:

Arrange numerator in the following manner:

\(\displaystyle \displaystyle\int \dfrac{5x - 2}{\sqrt{x^{2} + 2x}} dx = \tfrac{5}{2} \int \dfrac{2x - \frac{4}{5}}{\sqrt{x^{2} + 2x}}\,dx\) . What's going on here?

Let \(\displaystyle u \,=\, x^2+2x \quad\Rightarrow\quad du \,=\,(2x+2)dx \)

If we had \(\displaystyle 2x+2\) in the numerator, we're golden!
. . But we don't, do we?
Can we get it?

Yes, we can hammer the numerator into that form.


We have: .\(\displaystyle 5x-2\)

Multiply by \(\displaystyle \frac{5}{2}\!\cdot\!\frac{2}{5}\!:\;\;\frac{5}{2} \!\cdot\!\frac{2}{5}(5x-2) \:=\:\frac{5}{2}\left(2x - \tfrac{4}{5}\right)\)

Add and subtract 2: .\(\displaystyle \frac{5}{2}\left(2x \color{red}{+ 2} -\tfrac{4}{5} \color{red}{- 2}\right) \:=\:\frac{5}{2}\left(2x + 2 - \tfrac{14}{5}\right)\)

And we have: .\(\displaystyle \frac{5}{2}(2x+2) - 7\)


The integral becomes: .\(\displaystyle \displaystyle \int\frac{\frac{5}{2}(2x+2) - 7}{(x^2+2x)^{\frac{1}{2}}}\,dx \)

. . . . . . . . . . . . . . \(\displaystyle \displaystyle=\;\tfrac{5}{2}\int\frac{2x+2}{(x^2+2x)^{\frac{1}{2}}}\,dx \;-\; 7\int \frac{dx}{(x^2+2x)^{\frac{1}{2}}} \)
Got it?

 

[Jason76]

Thanks for the help. At least headed in the right direction.