# Substitution Indefinite Integral Problem- Just need work checked

• Apr 10th 2009, 11:04 PM
Jim Marnell
Substitution Indefinite Integral Problem- Just need work checked
Not sure if i've done this problem right.

$\displaystyle \int (\frac{6x^2}{(2x^3+7)^{3/2}})dx$

$\displaystyle u=2x^3+7$

$\displaystyle du=6x^2$

$\displaystyle \int (\frac{6x^2}{(2x^3+7)^{3/2}})dx$

$\displaystyle \int (\frac{du}{u^{3/2}})$

$\displaystyle \int (2u^{-3/2}+c$

$\displaystyle 2(2x^3+7)^{-3/2}+c$

$\displaystyle (4x^3+14)^{-3/2}+c$
• Apr 10th 2009, 11:11 PM
bebrave
Yeah you are right i think. Because i solved and i found same (Wink)
• Apr 10th 2009, 11:21 PM
mr fantastic
Quote:

Originally Posted by Jim Marnell
Not sure if i've done this problem right.

$\displaystyle \int (\frac{6x^2}{(2x^3+7)^{3/2}})dx$

$\displaystyle u=2x^3+7$

$\displaystyle du=6x^2 \, {\color{red}dx}$ Mr F says: Note the red stuff.

$\displaystyle \int (\frac{6x^2}{(2x^3+7)^{3/2}})dx$

$\displaystyle \int (\frac{du}{u^{3/2}})$

$\displaystyle \int (2u^{-3/2}+c$ Mr F says: Wrong from here onwards. See below.

$\displaystyle 2(2x^3+7)^{-3/2}+c$

$\displaystyle (4x^3+14)^{-3/2}+c$

$\displaystyle \int\! \frac{du}{u^{3/2}} = \int u^{-3/2} \, du = -2 u^{-1/2} + C$ etc.

Quote:

Originally Posted by bebrave
Yeah you are right i think. Because i solved and i found same

Then you're wrong too.
• Apr 10th 2009, 11:28 PM
bebrave
Thank you Mr.Fantastic...
Quote:

Yeah you are right i think. Because i solved and i found same
• Apr 11th 2009, 05:40 AM
Jim Marnell

$\displaystyle (-4x^3-14)^{-1/2}+c$
• Apr 11th 2009, 02:51 PM
mr fantastic
Quote:

Originally Posted by Jim Marnell
$\displaystyle (-4x^3-14)^{-1/2}+c$
When you substitute $\displaystyle u = 2x^3 + 7$ back into $\displaystyle -2 u^{-1/2} + C$ you do NOT get $\displaystyle (-4x^3-14)^{-1/2}+C$. The -2 does NOT and CANNOT get taken inside the brackets.
By the way, even if the answer was $\displaystyle 2 u^{-1/2} + C$, this is NOT equal to $\displaystyle (4x^3+14)^{-1/2}+C$.