what is the integral x(2x+1)^0.5 dx?

Question

.sam. · Answer

$$\int\limits x \sqrt{2 x+1} \, dx

u=2x+1 , du=2dx
du/2=dx
\[\text{}=\frac{1}{4}\int\limits (u-1) \sqrt{u} \, du$$

$$=\frac{1}{4}\int\limits \left(u^{3/2}-\sqrt{u}\right) \, du$$

$$\frac{1}{4}\int\limits u^{3/2} \, du-\frac{1}{4}\int\limits \sqrt{u} \, du$$

$$\text{}=\frac{1}{4}\int\limits u^{3/2} \, du-\frac{u^{3/2}}{6}$$

$$=\frac{u^{5/2}}{10}-\frac{u^{3/2}}{6}+\text{constant}$$

$$\text{}=\frac{1}{10} (2 x+1)^{5/2}-\frac{1}{6} (2 x+1)^{3/2}+\text{constant}$$

$$\text{}=\frac{1}{15} (2 x+1)^{3/2} (3 x-1)+\text{constant}$$

anonymous · Answer

integral?  looks like a simple u-substitution.

lgbasallote · Answer

oh man...why did i derive? im so screwed up..sorry asker!

.sam. · Answer

Wait just now i typed ""u=2x+1 , du=2dx"" it disappeared?

lgbasallote · Answer

haha =)) the site's been having bugs these past few days..we shouldnt be surprised

anonymous · Answer

thank you guys, yo spend some time trying to figure out how do i get around this