When I work at this I get it as being 4[(x+2)^3/4 ] / 3 + C Is this correct?

Printable View

- Jun 4th 2008, 04:59 AMCrakaindefinate integral x/(x+2)^1/4
When I work at this I get it as being 4[(x+2)^3/4 ] / 3 + C Is this correct?

- Jun 4th 2008, 05:12 AMPaulRS
mmmm, no, you forgot something there

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

$\displaystyle \int(x+2)^{3/4}dx=\frac{4}{7}\cdot{(x+2)^{7/4}}+k_1$

$\displaystyle \int(x+2)^{-1/4}dx=\frac{4}{3}\cdot{(x+2)^{3/4}}+k_2$

Thus: $\displaystyle \int\frac{x}{(x+2)^{1/4}}dx=\frac{4}{7}\cdot{(x+2)^{7/4}}-\frac{8}{3}\cdot{(x+2)^{3/4}}+k$ - Jun 4th 2008, 05:21 AMCraka
Thanks PaulRS but how/why do you get x+2-2 on the numerator ?

- Jun 4th 2008, 06:09 AMmr fantastic
Note that x + 2 - 2 = x. It is done to facilitate breaking the integral up into two more obvious integrals. As an alternative, you could make the substitution u = x + 2 and go from there.

By the way, you can check an answer to an integal by differentiating it and seeing whether the result is equal to the integrand.