Substitution Indefinite Integral Problem- Just need work checked

**Jim Marnell** · April 10th 2009, 11:04 PM

Not sure if i've done this problem right.

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

$u=2x^3+7$

$du=6x^2$

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

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

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

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

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

**bebrave** · April 10th 2009, 11:11 PM

Yeah you are right i think. Because i solved and i found same

**mr fantastic** · April 10th 2009, 11:21 PM

Originally Posted by Jim Marnell

Not sure if i've done this problem right.

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

$u=2x^3+7$

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

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

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

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

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

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

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

Originally Posted by bebrave

Yeah you are right i think. Because i solved and i found same

Then you're wrong too.

**bebrave** · April 10th 2009, 11:28 PM

Thank you Mr.Fantastic...

Yeah you are right i think. Because i solved and i found same

**Jim Marnell** · April 11th 2009, 05:40 AM

so the answer instead would be:

$(-4x^3-14)^{-1/2}+c$

**mr fantastic** · April 11th 2009, 02:51 PM

Originally Posted by Jim Marnell

so the answer instead would be:

$(-4x^3-14)^{-1/2}+c$

No.

When you substitute $u = 2x^3 + 7$ back into $-2 u^{-1/2} + C$ you do NOT get $(-4x^3-14)^{-1/2}+C$ . The -2 does NOT and CANNOT get taken inside the brackets.

By the way, even if the answer was $2 u^{-1/2} + C$ , this is NOT equal to $(4x^3+14)^{-1/2}+C$ .

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