Integrate exponential function (two parts)

Hello
Please help me solve this integral.
$\int(4x + 6)e^{x^2+3x} dx$
The integral has two parts
$u = 4x+6$
$\dfrac{dv}{dx} = e^{x^2+3x}$
How can I integrate $e^{x^2+3x}$ when the constant $e$ is raised to an exponent that has x squared as a term as in $e^{x^2}$?
Thank you.

Hello
Please help me solve this integral.
$\int(4x + 6)e^{x^2+3x} dx$
The integral has two parts
$u = 4x+6$
$\dfrac{dv}{dx} = e^{x^2+3x}$
How can I integrate $e^{x^2+3x}$ when the constant $e$ is raised to an exponent that has x squared as a term as in $e^{x^2}$?
Thank you.

Do you notice anything about how the multiplier, u(x), is related to the power? If so, ...

Were you told to do parts?
It is the reverse chain rule. You can look this up in your year 2 textbook.

It would be helpful if you provided a step by step solution so I can see the logic for myself.

Sorry, the sticky at the top of the forum says that you shouldnt expect worked solutions. The aim is to give hints which allow you to solve it for yourself. The two previous replies pretty much tell you want to do and its almost a "write down" solution anyway.

Ok. Here is my working so far.
$\int(4x + 6)e^{x^2+3x} dx$
The integral has two parts
$u = 4x+6$
$\dfrac{dv}{dx} = e^{x^2+3x}$
$\dfrac{du}{dx}u = 4$
$\dfrac{dv}{dx} = e^{x^2+3x}$
Is this how I integrate $e^{x^2+3x}$
$\dfrac{dv}{dx} = e^{x^2+3x}$ $/dfrac{1}{(2x+3)} e^{x^2+3x}$
Is the bottom line of working correct? I used the general rule $\int e^{ax+b} dx = \dfrac{1}{(a)} e^{ax+b}$

Ok. Here is my working so far.
$\int(4x + 6)e^{x^2+3x} dx$
The integral has two parts
$u = 4x+6$
$\dfrac{dv}{dx} = e^{x^2+3x}$
$\dfrac{du}{dx}u = 4$
$\dfrac{dv}{dx} = e^{x^2+3x}$
Is this how I integrate $e^{x^2+3x}$
$\dfrac{dv}{dx} = e^{x^2+3x}$ $\dfrac{1}{(2x+3)} e^{x^2+3x}$
Is the bottom line of working correct? I used the general rule $\int e^{ax+b} dx = \dfrac{1}{(a)} e^{ax+b}$

The bottom line is incorrect. Your "general" rule is only valid when a is constant.
A rule that works a bit more generally: $\displaystyle \int g'(x) e^{g(x)} \,dx = e^{g(x)}$
Even more generally: if $\displaystyle \int f(x)\,dx = F(x)$, then $\displaystyle \int g'(x) f(g(x)) \,dx = F(g(x))$ (**).
An important point here is that rewriting (**) as $\displaystyle \int f(g(x)) \,dx = \dfrac{F(g(x))}{g'(x)}$ is (in general) invalid - you're basically assuming that $\displaystyle \int \dfrac{g'(x)f(g(x))}{g'(x))} \,dx= \dfrac{\int g'(x)f(g(x))\,dx}{g'(x)}$, which is false unless g'(x) is constant.

This is helpful but I have another question.
How would I integrate $e^{x^2}$
$\int e^{x^2+3x} dx$

This is helpful but I have another question.
How would I integrate $e^{x^2}$
$\int e^{x^2+3x} dx$

This is helpful but I have another question.
How would I integrate $e^{x^2}$

Assuming its still the question in the OP, to repeat the hint in the first reply, what do you notice about how the multiplier
(4x+6)
is related to the exponent (x^2 + 3x) in terms of calculus operations? If you spot that, the integral is pretty much a write down.

Yes I notice that $4x+6$ could be the derivative of
$e^{x^2+3x}$
Can you proceed with the working.

Yes I notice that $4x+6$ could be the derivative of
$e^{x^2+3x}$
Can you proceed with the working.