1. Integrate

I am stuck on these three problems because I didn't understand how the teacher was teaching it. The test is in two days.

1.∫(x^3-2x+2)dx

2.∫(sqrt(x) - 1/x^2)dx

3.∫(2x^3+1)^2 x^2 dx

Thanks for the help.

2. Originally Posted by uniquereason81
I am stuck on these three problems because I didn't understand how the teacher was teaching it. The test is in two days.

1.∫(x^3-2x+2)dx

2.∫(sqrt(x) - 1/x^2)dx

3.∫(2x^3+1)^2 x^2 dx

Thanks for the help.
1. $\int{x^3-2x+3}dx=\frac{x^4}{4}-x^2+2x+C$...you do this by basic integration technique...
2. $\int{\sqrt{x}-\frac{1}{x^2}dx}=\int{x^{\frac{1}{2}}-x^{-2}dx}=\frac{2x^{\frac{3}{2}}}{3}+\frac{1}{x}+C$
3. $\int{(2x^3+1)^2\cdot{x^2}}dx=\frac{1}{6}\int{6x^2\ cdot(2x^3+1)^2dx}=\frac{(2x^3+1)^3}{18}+C$

3. Just so you know

Originally Posted by Mathstud28
1. $\int{x^3-2x+3}dx=\frac{x^4}{4}-x^2+2x+C$...you do this by basic integration technique...
2. $\int{\sqrt{x}-\frac{1}{x^2}dx}=\int{x^{\frac{1}{2}}-x^{-2}dx}=\frac{2x^{\frac{3}{2}}}{3}+\frac{1}{x}+C$
3. $\int{(2x^3+1)^2\cdot{x^2}}dx=\frac{1}{6}\int{6x^2\ cdot(2x^3+1)^2dx}=\frac{(2x^3+1)^3}{18}+C$
the second one I just converted the variables to a more hospitable form and applied general integration technique..

the second one I just made it so it was an example of $\int{f(g(x))\cdot{g'(x)}}dx$ after doing this I could integrate $f(g(x))$ as though it was just $f(x)$...so by making the derivative of the inside on the outside I could integrate $(2x^3+1)^2$ as though it was just $x^2$ and get $x^3$ and then you substitute back in the $g(x)$

4. wow i wish you could be my math teacher

5. Haha

Originally Posted by uniquereason81
wow i wish you could be my math teacher
Considering I am only in eleventh grade I dont think thats going to happen..but anytime you need help...Im here!

6. LOL you should think about teaching as a side job. Making some big bucks.

7. Haha

Originally Posted by uniquereason81
LOL you should think about teaching as a side job. Making some big bucks.
Someday my friend...maybe I'll be your kid's college professor

8. Hey man these three are giving me problems

1) ∫(2x^3 + )^2 x*dx

2) ∫x*sqrt(2x^3 + 1) dx

3) ∫ (x + 1)dx/cube root(x^2 + 2x -2)

Number 3 is the most confusing to me.
THANKS

9. Next time start a new thread

Originally Posted by uniquereason81
Hey man these three are giving me problems

1) ∫(2x^3 + )^2 x*dx

2) ∫x*sqrt(2x^3 + 1) dx

3) ∫ (x + 1)dx/cube root(x^2 + 2x -2)

Number 3 is the most confusing to me.
THANKS
You need to rewrite these....they dont make sense..

10. oops sorry.

1) ∫(2x^3 + 1)^2 * x * dx

2) ∫ x square root (2x^3 + 1 ) dx

3) ∫ (x +1 ) dx over cube root of (x^2 + 2x -2)

i don't know how to symbols

11. IT is ok

Originally Posted by uniquereason81
oops sorry.

1) ∫(2x^3 + 1)^2 * x * dx

2) ∫ x square root (2x^3 + 1 ) dx

3) ∫ (x +1 ) dx over cube root of (x^2 + 2x -2)

i don't know how to symbols

http://www.mathhelpforum.com/math-he...-tutorial.html first of all that will teach you how to use LaTeX and be better understood...now on to the integrals $\int{(2x^3+1)^2\cdot{x}dx}$...for this one honestly I would jsut distribute and integrate...or use the method I will show you on the others

2) $\int{x\sqrt{2x^3+1}dx}$..for this you must make a u-substitution..so you say that $u=\sqrt{2x^3+1}$...then say that $\sqrt[{3}]{\frac{u^2-1}{2}}=x$..and finally then you have that $dx=\frac{\sqrt[{3}]{4}u}{2(u^2-1)^{\frac{2}{3}}}$..and now you can see that this has no means that you could use to find the integral...there are ways but trust me for your level you must have copied it down wrong

3)Ok for this one we have $\int{(x+1)\cdot[x^2+2x-2}]^{\frac{-1}{3}}dx}$which I wont do for you but I will give you a hint $\frac{1}{2}\int{(2x+2)\cdot[x^2+2x-2]^{\frac{-1}{3}}dx}$

12. I got this stuff down pact. And thanks for the website.

13. Hey

Originally Posted by uniquereason81
I got this stuff down pact. And thanks for the website.
No problem anytime