Find the integral of: sqrt(2x-1)/(2x+3) dx

[MarkSA]

Hello,

1) Find the integral of: sqrt(2x - 1)/(2x + 3) dx

I'm not seeing anything in my integration steps that would work for this except integration by parts perhaps?

Let u = 2x + 3
du = 2xdx
Let dv = (2x - 1)^(1/2)
v = 1/3 * (2x - 1)^(3/2)

= (2x + 3)* 1/3 * (2x - 1)^(3/2) - 1/2 integral of: (2dx)[1/3 * (2x - 1)^(3/2)]

This seems to get me.. (2x + 3)(1/3 * (2x - 1)^(3/2)) - 2/15 * (2x - 1)^(5/2) + C

But it's conflicting with the book's answer again so i'm not sure if I did it correctly.. plugging in a number randomly doesn't give the same results or the same difference in results for the two answers.

The book's answer is: sqrt(2x - 1) - 2arctan(sqrt(2x - 1)/2) + C

 

[o_O]

\(\displaystyle \int udv = uv - \int v du\)

Using your substitutions, you would get:

\(\displaystyle \int u dv = \int (2x + 3)(2x - 1)^{\frac{1}{2}}dx \neq \int \sqrt{\frac{2x -1}{2x+3}}dx\)

Also, if u = 2x + 3, then du = 2dx.

 

[soroban]

Hello, Mark!

Here's a substitution I saw years ago . . .

\(\displaystyle \text{1) Integrate: }\;\int\sqrt{\frac{2x - 1}{2x + 3}}\, dx\)

\(\displaystyle \text{Let: }\;u \;=\;\sqrt{\frac{2x-1}{2x+3}}\quad\Rightarrow\quad u^2 \:=\:\frac{2x-1}{2x+3}\quad\Rightarrow\quad 2u^2x + 3u^2\:=\:2x-1\)

. . . . \(\displaystyle 2x-2u^2x \;=\;3u^2-1\quad\Rightarrow\quad 2x(1-u^2) \;=\;3u^2-1\quad\Rightarrow\quad x \;=\;\frac{3u^2-1}{2(1-u^2)}\)

. . . . \(\displaystyle \text{Hence: }\;dx\;=\;\frac{2u\,du}{(1-u^2)^2}\)


\(\displaystyle \text{Substitute: }\;\int u\cdot\frac{2u\,du}{(1-u^2)^2} \;=\;\int\frac{2u^2}{(1-u^2)^2}\,du\)


I see two approaches now . . . neither is advised . . .LOL!



\(\displaystyle \text{[1] Trig Substitution}\)

\(\displaystyle \text{Let }u \,=\,\sin\theta\quad\Rightarrow\quad du \,=\,\cos\theta\,d\theta\quad\Rightarrow\quad 1 - u^2 \:=\:\cos^2\theta\)

\(\displaystyle \text{Substitute: }\:\int\frac{2\sin^2\!\theta}{\cos^4\!\theta}(\cos\theta\,d\theta) \;=\;2\int\frac{\sin^2\!\theta}{\cos^3\!\theta}\,d\theta \;=\;2\int\frac{1}{\cos\theta}\cdot\frac{\sin^2\!\theta}{\cos^2\!\theta}\,d\theta \;=\;2\int\sec\theta\tan^2\!\theta\,d\theta\)

\(\displaystyle \text{This is an unpleasant integral: }\;2\int\sec\theta(\sec^2\!\theta - 1)\,d\theta \;=\;2\int\left(\sec^3\!\theta - \sec\theta)\,d\theta\)

\(\displaystyle \text{If you happen to }know\text{ the integral of secant-cubed, it isn't difficult.}\)
. . \(\displaystyle \text{If not, some Integration-by-parts is in order.}\)



\(\displaystyle \text{[2] Partial Fractions}\)

\(\displaystyle \text{Since }(1 - u^2)^2 \;=\;[(1-u)(1+u)]^2\;=\;(1-u)^2(1+u)^2\)

. . \(\displaystyle \text{we have: }\;\frac{2u^2}{(1-u)^2(1+u)^2} \;=\;\frac{A}{1-u} + \frac{B}{(1-u)^2} + \frac{C}{1+u} + \frac{D}{(1+u)^2}\)

Bon voyage!

 

[Deleted member 4993]

Hello,

1) Find the integral of: sqrt(2x - 1)/(2x + 3) dx

I'm not seeing anything in my integration steps that would work for this except integration by parts perhaps?

Let u = 2x + 3
du = 2xdx
Let dv = (2x - 1)^(1/2)
v = 1/3 * (2x - 1)^(3/2)

= (2x + 3)* 1/3 * (2x - 1)^(3/2) - 1/2 integral of: (2dx)[1/3 * (2x - 1)^(3/2)]

This seems to get me.. (2x + 3)(1/3 * (2x - 1)^(3/2)) - 2/15 * (2x - 1)^(5/2) + C

But it's conflicting with the book's answer again so i'm not sure if I did it correctly.. plugging in a number randomly doesn't give the same results or the same difference in results for the two answers.

The book's answer is: sqrt(2x - 1) - 2arctan(sqrt(2x - 1)/2) + C

Here the idea should be to get rid of the \(\displaystyle \sqrt\) sign.

so substitute

\(\displaystyle \sqrt{2x-1} = u\)

\(\displaystyle \frac{dx}{\sqrt{2x-1}} = du\)

\(\displaystyle dx = u\cdot du\)

then

\(\displaystyle \int{\frac{\sqrt{2x-1}}{2x+3}}dx\)

\(\displaystyle = \int{\frac{u}{u^2+4}u}du\)

\(\displaystyle = \int du - \int\frac{4}{u^2+4}}du\)

Now finish it....