Intergal question

[markraz]

Hi is this the way to go on this??

\(\displaystyle \Large\int \frac{x^2}{(x+1)^4} dx\)

\(\displaystyle \Large u = (x+1)\)

\(\displaystyle \Large\int \frac{x^2}{(u)^4} dx\)

\(\displaystyle \Large du = 1\)
\(\displaystyle \Large dx = du\)

\(\displaystyle \Large\int \frac{x^2}{(u)^4} du\)

I think this is correct up to here:

but is this the way to proceed??
\(\displaystyle \Large x = (u-1)\)

\(\displaystyle \Large\int \frac{(u-1)^2}{(u)^4} du\)

if so do I FOIL? this out \(\displaystyle (u^2-2u+1)\)

\(\displaystyle \Large\int \frac{(u^2-2u+1)}{(u)^4} du\)

then ???
\(\displaystyle \Large\int \frac{u^2}{u^4}- \frac{2u}{u^4} + \frac{1}{u^4}du\)

thanks

 

[stapel]

is this the way to proceed??
\(\displaystyle \Large x = (u-1)\)

\(\displaystyle \Large\int \frac{(u-1)^2}{(u)^4} du\)

if so do I FOIL? this out \(\displaystyle (u^2-2u+1)\)

\(\displaystyle \Large\int \frac{(u^2-2u+1)}{(u)^4} du\)

then ???
\(\displaystyle \Large\int \frac{u^2}{u^4}- \frac{2u}{u^4} + \frac{1}{u^4}du\)

Thank you for showing your steps and reasoning so clearly!

Yes, what you've done is how I would have proceeded. Now restate the fractions:

. . . . .\(\displaystyle \dfrac{u^2}{u^4}\, -\, \dfrac{2u}{u^4}\, +\, \dfrac{1}{u^4}\, =\, \dfrac{1}{u^2}\, -\, \dfrac{2}{u^3}\, +\, u^{-4}\, =\, u^{-2}\, -\, 2u^{-3}\, +\, u^{-4}\)

Now use the Power Rule to integrate term-by-term.

 

[markraz]

Thank you for showing your steps and reasoning so clearly!

Yes, what you've done is how I would have proceeded. Now restate the fractions:

. . . . .\(\displaystyle \dfrac{u^2}{u^4}\, -\, \dfrac{2u}{u^4}\, +\, \dfrac{1}{u^4}\, =\, \dfrac{1}{u^2}\, -\, \dfrac{2}{u^3}\, +\, u^{-4}\, =\, u^{-2}\, -\, 2u^{-3}\, +\, u^{-4}\)

Now use the Power Rule to integrate term-by-term.

thanks, I originally did this out by hand but , then when I plugged it into Wolfram and MathCAD to check my answer, I got 2 different answers , maybe I did the power rule wrong??
anyway..Thanks for your help.

 

[stapel]

thanks, I originally did this out by hand but , then when I plugged it into Wolfram and MathCAD to check my answer, I got 2 different answers , maybe I did the power rule wrong??

Since we can't see what you entered into Wolfram Alpha or MathCAD, don't know what results you obtained, and can't see your steps, I'm afraid there is no way to answer this. Sorry.