Question

∫16x/(8x^2+2)

Answer 1

\[\large \int\limits\frac{\color{royalblue}{16x\;dx}}{\color{orangered}{8x^2+2}}\]

Letting \(\large \color{orangered}{u=8x^2+2}\)

What do we get if we take the derivative of our substitution? :o

Answer 2

16x

Answer 3

\[\large \color{royalblue}{du=16x\;dx}\]Yes, good! :)
I tried to color-code them so you can see how we're going to make our substitution.

Answer 4

\[\large \int\limits\limits\frac{\color{royalblue}{16x\;dx}}{\color{orangered}{8x^2+2}} \qquad\to\qquad \int\limits\limits\frac{\color{royalblue}{du}}{\color{orangered}{u}}\]
Understand how we did that? :O
That should make our integral easier to evaluate.
We can write it like this if it makes it easier to read.\[\large \int\limits \frac{1}{u}\;du\]

Answer 5

Yes, so the intergral is lnu?

Answer 6

Yes, good.
As a final step, rewrite your answer in terms of x.
Meaning ~ Undo your substitution now.

Answer 7

\[\ln \left| 8x ^{2}+2 \right|+c\]

Answer 8

Cool, good job! \c:/