# Indefinite Integral Problem- Can anyone check my work

• Apr 10th 2009, 08:57 PM
Jim Marnell
Indefinite Integral Problem- Can anyone check my work
$\int (\frac{2}{3x^4})dx$
$\int (2\times{3x^{-4}})dx$
$\int (6x^{-4})dx$
$6\int (x^{-4})dx$
$6 (\frac{x^{-3}}{-3})+c$

$-2x^{-3}+c$

$\frac{-2}{x^3}+c$

Thanks!
• Apr 10th 2009, 10:12 PM
Quote:

Originally Posted by Jim Marnell
$\int (\frac{2}{3x^4})dx$

$\int{ (2\times {\color{red}3} x^{-4}})dx$

The three in first step was in denominator
the second step should be

$\frac{1}{3}\int (2\times{x^{-4}})dx$

Thanks!

Red
• Apr 10th 2009, 10:29 PM
Jim Marnell
$\int (\frac{2}{3x^4})dx$

$\frac{1}{3}\int (2\times{x^{-4}})dx$

$\frac{1}{3}\int (2x^{-4})dx$

$\frac{1}{3}\times{2}\int (x^{-4})dx$

$\frac{2}{3}\times{\frac{x^{-3}}{-3}}+c$

$\frac{-2}{9}x^{-3}+c$

$\frac{\frac{2}{9}}{x^3}+c$

I'm not sure if my final step is right. Thanks for any help!
• Apr 10th 2009, 10:38 PM
Quote:

Originally Posted by Jim Marnell
$\int (\frac{2}{3x^4})dx$

$\frac{1}{3}\int (2\times{x^{-4}})dx$

$\frac{1}{3}\int (2x^{-4})dx$

$\frac{1}{3}\times{2}\int (x^{-4})dx$

$\frac{2}{3}\times{\frac{x^{-3}}{-3}}+c$

$\frac{-2}{9}x^{-3}+c$

$\frac{\frac{{\color{red}-}2}{9}}{x^3}+c$

You forgot "-" sign , his can further be written as

$\frac{-2}{9x^3}+c$

I'm not sure if my final step is right. Thanks for any help!

Red

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