MHB What is the integral of $\frac{x^2-1}{\sqrt{2x-1}}$?

[karush]

$\tiny{307.4.5.71}$
\begin{align}
\displaystyle
I_{71}&=\int{\frac{x^2-1}{\sqrt{2x-1}}}dx\\
u&=2x-1\therefore dx=\frac{1}{2}du\\
x&=\frac{u+1}{2}\\
I_u&= \int{\frac{{[(1/2)(u+1)]}^2-1}{\sqrt{u}}}
\cdot \frac{1}{2} \, du\\
\end{align}

$\textit{not sure what is best to do next?}$

 

[MarkFL]

Your substitution is a good one...let's write the integral as:

$$I=\frac{1}{2}\int\frac{\left(\frac{u+1}{2}\right)^2-1}{u^{\frac{1}{2}}}\,du$$

Next, let's multiply by $$1=\frac{4}{4}$$:

$$I=\frac{1}{8}\int\frac{(u+1)^2-4}{u^{\frac{1}{2}}}\,du$$

Can you proceed?

 

[karush]

$\tiny{307.4.5.71}$
\begin{align}
\displaystyle
I_{71}&=\int{\frac{x^2-1}{\sqrt{2x-1}}}dx\\
u&=2x-1\therefore dx=\frac{1}{2}du\\
x&=\frac{u+1}{2}\\
I_u&= \int{\frac{{[(1/2)(u+1)]}^2-1}{\sqrt{u}}}
\cdot \frac{1}{2} \, du\\
&=\frac{1}{8}\int\frac{(u+1)^2-4}{u^{\frac{1}{2}}}\,du
=\frac{1}{8}\int\frac{u^2+2u-3}{u^{1/2}}
\end{align}

$\text{this or go with $v=u+1$ for a trig id}$

 

[MarkFL]

Just simplify the integrand...you will have 3 terms containing u to various powers, to which you can then apply the power rule. (Yes)

 

[Greg]

$$\int\frac{x^2-1}{\sqrt{2x-1}}\,\text{d}x$$

$$u=\sqrt{2x-1}\quad x=\frac{u^2+1}{2}$$

$$\text{d}u=\frac{1}{\sqrt{2x-1}}\text{ d}x$$

$$\int\left(\frac{u^2+1}{2}\right)^2-1\text{ d}u$$

 

[karush]

$\tiny{307.4.5.71}$
\begin{align}
\displaystyle
I_{71}&=\int{\frac{x^2-1}{\sqrt{2x-1}}}dx\\
u&=2x-1\therefore dx=\frac{1}{2}du\\
x&=\frac{u+1}{2}\\
I_u&= \int{\frac{{[(1/2)(u+1)]}^2-1}{\sqrt{u}}}
\cdot \frac{1}{2} \, du\\
&=\frac{1}{8}\int\frac{(u+1)^2-4}{u^{\frac{1}{2}}}\,du
=\frac{1}{8}\int\frac{u^2+2u-3}{u^{1/2}}\, du\\
&=\frac{1}{8}\left[\int\frac{u^2}{u^{1/2}} \, du
+2\int\frac{u}{u^{1/2}}\, du
-3\int\frac{1}{u^{1/2}}\, du\right] \\
&=\frac{1}{8}\left[\dfrac{2u^\frac{5}{2}}{5}
+\dfrac{4u^\frac{3}{2}}{3}
-6\sqrt{u}\right]\\
I&=\dfrac{\frac{\left(2x-1\right)^\frac{5}{2}}{5}+\frac{2\left(2x-1\right)^\frac{3}{2}}{3}+\sqrt{2x-1}}{4}-\sqrt{2x-1}
\end{align}