Question

Please help me finish solving the following integral (click to see).

Answer 1

\[\int\limits_{}^{}\frac{ x }{ \sqrt{9-x^4} }dx\]

Answer 2

\[u=x^2\]

Answer 3

\[\int\limits_{}^{}\frac{ x }{ \sqrt{9-u^2} }\frac{ du }{ 2x }\]

Answer 4

\[\frac{ 1 }{ 2 } \int\limits_{}^{}\frac{ 1 }{ \sqrt{9-u^2} }du\]

Answer 5

now i'm stuck

Answer 6

A way that you could do it is factor 9 out of the bottom so as to get\[1/2 \int\limits \frac{ 1 }{ 3\sqrt{1-\frac{ u^2 }{ 9 }} }\] and then you factor the 1/3 outside of the integral to get \[1/6 \int\limits \frac{ 1 }{ \sqrt{1-\frac{ u^2 }{ 9 }} }\] Can you go from there?

Answer 7

HINT: \(\dfrac{d}{dx} \sin^{-1}x = \dfrac{1}{\sqrt{1-x^2}}\)