Question

How would you solve...

Answer 1

?

Answer 2

lol

Answer 3

\[\int\limits_{}^{}\frac{ dx }{ \sqrt{3-x}-\sqrt{(3-x)^3} }\]

Answer 4

sorry, i was writing the integral xD

Answer 5

use $$t=\sqrt{3-x}$$ then, $$dt=\frac{1}{2\sqrt{3-x}}dx$$
$$\begin{align*}
\int \frac{dx}{\sqrt{3-x}-\sqrt{3-x}^3}&=2\int \frac{1}{1-\sqrt{3-x}^2}\frac{dx}{2\sqrt{3-x}}\\
&=2\int \frac{1}{1-t^2}dt
\end{align*}$$

Answer 6

I'm sorry, can you explain how did you turn \[\sqrt{3-x}\] into a 1? Also, i think you forgot a - sign in the dt

Answer 7

yeah it should be $$dt=\frac{-1}{2\sqrt{3-x}}dx$$
Ok I'll go step by step then,
$$\begin{align*}
\int \frac{1}{\sqrt{3-x}-\sqrt{3-x}^3 }dx&=\int \frac{1}{\sqrt{3-x}(1-\sqrt{3-x}^2)}dx\\
&=\int \frac{2}{2\sqrt{3-x}\left(1-\sqrt{3-x}^2\right)}dx\\
&=-2\int \frac{1}{1-\sqrt{3-x}^2}\frac{dx}{-2\sqrt{3-x}}\\\end{align*}$$

Answer 8

Oh, now i get it! Thanks a lot!

Hey, after staring at it for a while, I found a second way to solve it, look:

\[u^2=3-x\]
\[-dx=2udu => dx=-2udu\]

\[-2\int\limits_{}^{}\frac{ udu }{ \sqrt{u^2}-\sqrt{(u^2)^3} }\]
\[-2\int\limits_{}^{}\frac{ udu }{ u-u^3 }=-2\int\limits_{}^{}\frac{ du }{ 1-u^2 }\]

What do u think? Thanks a lot!

Answer 9

substitution is quite similar $$u^2=3-x\implies u=\sqrt{3-x}$$