# Help with this integral?

1. Jul 25, 2011

### 1MileCrash

1. The problem statement, all variables and given/known data

$\int (x^{2})(3^{x^{3}}) dx$

2. Relevant equations

3. The attempt at a solution

let u = x^3, du = 3x^2 dx

$\frac{1}{3}\int 3^{u}du$

$\frac{1}{3} (\frac{1}{ln 3})3^{u}$

$\frac{1}{3} (\frac{1}{ln 3})3^{x^{3}}$

2. Jul 25, 2011

### BruceW

Yep. You got it right.

3. Jul 25, 2011

### 1MileCrash

*Sigh* Mathway never agrees with me, or says it can't solve the problem. Wolfram also showed me something weird.

I guess I need to stop second guessing myself when those online solvers give a weird answer.

4. Jul 25, 2011

### dextercioby

Don't forget the constant of integration, each time you use the integral sign without limits. Wolfram is 99.99% right.

5. Jul 25, 2011

### SammyS

Staff Emeritus
(1/3) = 3-1

Therefore, $\displaystyle \frac{1}{3} \left(\frac{1}{\ln 3}\right)3^{x^{3}}=\left(\frac{1}{\ln 3}\right)3^{(x^{3}-1)}\,,$ which is pretty much what WolframAlpha gives.

6. Jul 25, 2011

### 1MileCrash

I think that was wolfram's result. Now I see why.

7. Jul 25, 2011

### gb7nash

Wolfram is correct. It gives me:

$$\frac{3^{x^3-1}}{\log{3}}$$

which is:

$$\frac{3^{x^3}3^{-1}}{\log{3}} = \frac{1}{3}\frac{3^{x^3}}{\log{3}}$$

Sammy beat me.

8. Jul 25, 2011

### ArcanaNoir

Whenever my calculator tells me it can't solve an integral I always try making some substitution or similar adjustments (especially trig). Sometimes the ability to see pieces of a puzzle is lacking in straight-up algorithms.