What is the integral of e^{x^2}x-e^{3x^4}12x^3 ?

The integral of e^{x^2}x-e^{3x^4}12x^3 is 1/2 e^{x^2}-e^{3x^4}+C

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integral of e^{x^2}x-e^{3x^4}12x^3

\int e^{x^{2}} x - e^{3 x^{4}} 12 x^{3} d x

\frac{1}{2} e^{x^{2}} - e^{3 x^{4}} + C

Solution steps

\int e^{x^{2}} x - e^{3 x^{4}} \cdot 12 x^{3} d x

Apply the Sum Rule:

\int f (x) \pm g (x) d x = \int f (x) d x \pm \int g (x) d x

= \int e^{x^{2}} x d x - \int e^{3 x^{4}} \cdot 12 x^{3} d x

\int e^{x^{2}} x d x = \frac{1}{2} e^{x^{2}}

\int e^{3 x^{4}} \cdot 12 x^{3} d x = e^{3 x^{4}}

= \frac{1}{2} e^{x^{2}} - e^{3 x^{4}}

Add a constant to the solution

= \frac{1}{2} e^{x^{2}} - e^{3 x^{4}} + C

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What is the integral of e^{x^2}x-e^{3x^4}12x^3 ?
The integral of e^{x^2}x-e^{3x^4}12x^3 is 1/2 e^{x^2}-e^{3x^4}+C