Integrating x² times 3^(x³)

[1MileCrash]

Homework Statement

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

Homework Equations

The Attempt at a Solution

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

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

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

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

 

[BruceW]

Yep. You got it right.

 

[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.

 

[dextercioby]

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

 

[SammyS]

(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.

 

[1MileCrash]

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

 

[gb7nash]

Wolfram also showed me something weird.

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.

 

[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.