Thanks! Glad you found it helpful.

[karush]

Whitman 8.3.12
$$\int \frac{{x}^{3 }}{\sqrt{4x ^2 - 1}} \ dx =
\frac{\left(2{x}^{2}+1\right)\sqrt{4{x}^{2}-1}}{24}$$
$$u=4x^2 - 1 \ \ \ \ du=8x \ dx \ \ \ x=\left(\frac{u-1}{4 }\right)^\frac{1}{2}$$
Substitute and simplify
$$\frac{1}{32}\displaystyle \int\dfrac{u+1}{\sqrt{u}}\,\mathrm{d}u$$

 

[Prove It]

Re: Whitman 8.3.12 \int \frac{{x}^{3 }}{\sqrt{4x ^2 - 1}} \ dx =

Whitman 8.3.12
$$\int \frac{{x}^{3 }}{\sqrt{4x ^2 - 1}} \ dx =
\frac{\left(2{x}^{2}+1\right)\sqrt{4{x}^{2}-1}}{24}$$
$$u=4x^2 - 1 \ \ \ \ du=8x \ dx \ \ \ x=\left(\frac{u-1}{4 }\right)^\frac{1}{2}$$
Substitute and simplify
$$\frac{1}{32}\displaystyle \int\dfrac{u+1}{\sqrt{u}}\,\mathrm{d}u$$

Actually it's $\displaystyle \begin{align*} x = \left( \frac{u + 1}{4} \right) ^{\frac{1}{2}} \end{align*}$, but it appears that you had this right anyway as you have gotten the correct integrand.

So now write $\displaystyle \begin{align*} \frac{1}{32} \int{ \frac{u + 1}{\sqrt{u}}\,\mathrm{d}u } = \frac{1}{32}\int{ \left( u^{\frac{1}{2}} + u^{-\frac{1}{2}} \right) \,\mathrm{d}u } \end{align*}$ and integrate.

 

[karush]

Yeah didn't copy my notes to good soooo..
$$\frac{1} {32}\left[\frac{2{u}^{3/2}}{3 }+2u^{1/2}\right]
=\left[\frac{\sqrt{u}\left(u+3\right)}{48}\right]$$
Back substittute $u=4{x}^{2}-$1 then

$I =
\frac{\left(2{x}^{2}+1\right)\sqrt{4{x}^{2}-1}}{24}$

 

[Prove It]

yeah didn't copy my notes to good soooo..
$$\frac{1} {32}\left[\frac{2{u}^{3/2}}{3 }+2u^{1/2}\right]
=\left[\frac{\sqrt{u}\left(u+3\right)}{48}\right]$$
back substittute $u=4{x}^{2}-$1 then

$i =
\frac{\left(2{x}^{2}+1\right)\sqrt{4{x}^{2}-1}}{24}$

+c :p

 

[MarkFL]

Let's try IBP...we are given:

$$I=\int \frac{x^3}{\sqrt{4x^2-1}},dx$$

Let:

$$u=x^2\implies du=2x\,dx$$

$$dv=\frac{1}{8}\cdot\frac{8x}{\sqrt{4x^2-1}}\,dv\implies v=\frac{1}{4}\sqrt{4x^2-1}$$

Hence:

$$I=\frac{x^2}{4}\sqrt{4x^2-1}-\frac{1}{16}\int 8x\sqrt{4x^2-1}\,dx$$

$$I=\frac{x^2}{4}\sqrt{4x^2-1}-\frac{1}{24}\left(4x^2-1\right)^{\frac{3}{2}}+C$$

$$I=\frac{\sqrt{4x^2-1}}{24}\left(6x^2-\left(4x^2-1\right)\right)+C$$

$$I=\frac{\sqrt{4x^2-1}}{24}\left(2x^2+1\right)+C$$

 

[karush]

Where does $dv \ \ \ \ \ $ come from?

 

[MarkFL]

Where does $dv \ \ \ \ \ $ come from?

The original integral may be written as:

$$I=\int u\,dv$$

 

[karush]

Got it

Liked the way that was solved