Integration by substitution

[griffis2]

Hi there,

[h=1]I'm really having difficulty finding the integral of this (3/x^3)(7/9x^2)^(3/2) dx

Any help at all would be appreciated. Thanks![/h]

 

[stapel]

I'm really having difficulty finding the integral of this (3/x^3)(7/9x^2)^(3/2) dx

I will guess, from the subject line, that the instructions told you to do something like "integrate using u-substitution". (Subject lines frequently don't relate, so apologies in advance if this is incorrect.) So you have:

. . . . .\(\displaystyle \mbox{Integrate: }\, \)\(\displaystyle \displaystyle \int\, \)\(\displaystyle \left(\dfrac{3}{x^3}\right)\, \left(\dfrac{7}{9x^2}\right)^{\frac{3}{2}}\, dx\)

You've got two fractions, so maybe you can express one in terms of the other. I would note the following:

. . . . .\(\displaystyle \left(9x^2\right)^{\frac{3}{2}}\, =\, \left(\sqrt{9x^2}\right)^3\, =\, \left(3 \bigg|\, x\, \bigg|\right)^3\, =\, 27\, \bigg|\, x^3\, \bigg|\)

Then:

. . . . .\(\displaystyle \left(\dfrac{7}{9x^2}\right)^{\frac{3}{2}}\, =\, \dfrac{7^{\frac{3}{2}}}{27\, \bigg|\, x^3\, \bigg|}\, =\, \left(\dfrac{7^{\frac{3}{2}}}{81}\right)\, \left(\dfrac{3}{\bigg|\, x^3\, \bigg|}\right)\)

Are you given any limitations on the values of the variable, so that the absolute-value bars can be dropped? Thank you!