how to calculate this indefinite integral?

Question

I'm studying for an exam and I found this excercise in my textbook, but I don't know how to evaluate this integral (I've never seen one with exponentials like that), any hint?

$\displaystyle\int 2^{x^3-x^2}(6x^2-4x)$

I noticed that part of the derivative of $2^{x^3-x^2}$ differs only by a constant from $(6x^2-4x)$ but I don't see how can I use this fact to evaluate the integral.

Answer 1

Hint:

$(6x^2-4x)dx=2d(x^3-x^2)$

so try the substitution $u=x^3-x^2$

Answer 2

$$x^3-x^2=u\implies (3x^2-2x)dx=du$$

$$\int2^{x^3-x^2}(6x^2-4x)dx=\int2^u\cdot2du=\cdots$$

Now, $\displaystyle\frac{d(a^x)}{dx}=a^x\cdot\ln a$ for real $a>0$

$\displaystyle\implies\int a^x\ dx=\frac{a^x}{\ln a}+K $

Answer 3

$(6x^2-4x)dx=2(3x^2-2x)dx=2d(x^3-x^2)$ and than we have:

$\displaystyle\int 2^{x^3-x^2}(6x^2-4x)dx=\displaystyle\int 2^{x^3-x^2}2d(x^3-x^2)=2\displaystyle\int 2^{x^3-x^2}d(x^3-x^2)=2\displaystyle\int 2^tdt$

Answer 4

What you have noted is exactly what you want to use with substitution.

Let $u = x^3 - x^2$. Then $du = 3x^2 - 2x$ meaning that $2du = (6x^2 - 4x) \; dx$.

So you get $$ \int 2^u 2 \; du. $$