Question

integrate (sqrt((x-1)/x^5)) using substitution (no trig substitution)

Answer 1

Trig substitution is not allowed, but I suppose I can do this:
\[\int \frac{\sqrt{x-1}}{x^5} \text{d}x\\
\begin{align}
\text{Let }& u =\sqrt{x-1}\\
&u^2=x-1\\
&2u du=dx\\
&(u^2+1)^5=x^5
\end{align}\]
Thus the integral becomes:
\[\int \frac{2u^3}{(u^2+1)^5}\text{d}u\]
Now let v=u^2+1.

Answer 2

wait wait i think u got the question wrongly. Its \[\int\limits\sqrt{x-1\div x^{5}}\]

Answer 3

\[\int\limits_{}^{}\frac{1}{x^2} \sqrt{\frac{x-1}{x} } dx\]

Answer 4

\[\int\limits_{}^{}\frac{1}{x^2} \sqrt{1-\frac{1}{x}} dx\]

Answer 5

let u be the inside of that square root thingy

Answer 6

okay i got it until there, and then?

Answer 7

\[u=1-\frac{1}{x} => du=\frac{1}{x^2} dx\]

Answer 8

\[\int\limits_{}^{}\sqrt{u} du\]

Answer 9

You can integrate that expression?

Answer 10

yessss. thank you! that helps alot! :)