Question

∫16x/(√8x^2+1)

Answer 1

\[\bf \int\limits_{}^{}\frac{ 16x }{ \sqrt{8x^2+1} }dx\]Is this what you have? @Aoi

Answer 2

Yes

Answer 3

Ok then. First factor out the 16 from the integral. Then make the following u-substitution:\[\bf u=8x^2+1 \implies \frac{du}{dx}=16x \implies dx = \frac{du}{16x}\]Then you get the following integral:\[\bf 16\int\limits_{}^{}\frac{ \cancel{x} }{ \sqrt{u} } \left( \frac{ du }{ 16 \cancel{x} } \right)=\int\limits_{}^{}\frac{ 1 }{ \sqrt{u} }du=\int\limits_{}^{}u^{-\frac{1}{2}}\ du\]Can you finish integrating that? After you're done, substitute \(\bf 8x^2+1\) back in for \(\bf u\) at the end. @Aoi

Answer 4

???? @Aoi u there?

Answer 5

Yes

Answer 6

Can you finish it off like I asked for above? @Aoi

Answer 7

Yes, thank you

Answer 8

Lol