What is the integral of 7x(e^{6x^2}-1) ?

The integral of 7x(e^{6x^2}-1) is 7(1/12 e^{6x^2}-(x^2)/2)+C

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integral of 7x(e^{6x^2}-1)

\int 7 x (e^{6 x^{2}} - 1) d x

7 (\frac{1}{12} e^{6 x^{2}} - \frac{x ^{2}}{2}) + C

Solution steps

\int 7 x (e^{6 x^{2}} - 1) d x

Take the constant out:

\int a \cdot f (x) d x = a \cdot \int f (x) d x

= 7 \cdot \int x (e^{6 x^{2}} - 1) d x

Expand

x (e^{6 x^{2}} - 1) : e^{6 x^{2}} x - x

= 7 \cdot \int e^{6 x^{2}} x - x 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

= 7 (\int e^{6 x^{2}} x d x - \int x d x)

\int e^{6 x^{2}} x d x = \frac{1}{12} e^{6 x^{2}}

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

= 7 (\frac{1}{12} e^{6 x^{2}} - \frac{x ^{2}}{2})

Add a constant to the solution

= 7 (\frac{1}{12} e^{6 x^{2}} - \frac{x ^{2}}{2}) + C

\frac{d}{d x} (10000 - 40 x - 0.02 x^{2})

\int \frac{3 x ^{2} + 14 x + 13}{x + 4} d x

\int \frac{1}{( x x - 1 )} d x

\int \frac{x ^{2} - 14 x - 5}{( x - 1 ) ^{2} ( x ^{2} + 1 )} d x

f (x) = 9 x + 5

What is the integral of 7x(e^{6x^2}-1) ?
The integral of 7x(e^{6x^2}-1) is 7(1/12 e^{6x^2}-(x^2)/2)+C