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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.

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  • $\begingroup$ Do you know how to find an antiderivative of $x\mapsto 2^x$? $\endgroup$ – Git Gud May 1 '14 at 14:48
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Hint:

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

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

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

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

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 $

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

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

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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 $$ \int 2^u 2 \; du. $$

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