UHWO s6.7.r.35 - Integral with substitutions

[karush]

$\large{s6.7.r.35}$

$$\displaystyle
I=\int\frac{1}{\sqrt{x+{x}^{3/2}}} \, dx $$

Not sure what to set $u$ to
online calculator. $u=\sqrt{x}$
But didn't look the best choice.

 

[Greg]

Re: UHWO s6.7.r.35

Hi karush. A condensed working:

\(\displaystyle \int\dfrac{1}{\sqrt{x+x^{3/2}}}\,dx\)

\(\displaystyle u^2=x,\quad2u\,du=\,dx\)

\(\displaystyle 2\int\dfrac{1}{\sqrt{1+u}}\,du\)

\(\displaystyle u=\sinh^2(w),\quad du=2\sinh(w)\cosh(w)\,dw\)

\(\displaystyle \begin{align*}4\int\sinh(w)\,dw&=4\cosh(w)+C \\
&=4\cosh\left(\sinh^{-1}(\sqrt u)\right)+C \\
&=4\cosh\left(\sinh^{-1}(x^{1/4})\right)+C \\
&=4\sqrt{1+\sqrt x}+C\end{align*}\)

 

[MarkFL]

Once you get to the point:

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

You can just go directly to (with the mental sub. $v=u+1$):

\(\displaystyle I=4(u+1)^{\frac{1}{2}}+C=4\sqrt{\sqrt{x}+1}+C\) :D

 

[karush]

well that sure beats some of other walk in woods solutions I saw

looks I need to get more familiar with \(\displaystyle sinh(x)\) and \(\displaystyle cosh(x)\)