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.

Reply 1

mqb2766

Original post

by eugenemcmoy

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, ...

Reply 2

MaxAOC

Original post

by eugenemcmoy

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.

Were you told to do parts?

It is the reverse chain rule. You can look this up in your year 2 textbook.

Reply 3

eugenemcmoy

Original post

by mqb2766

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

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

Reply 4

eugenemcmoy

Original post

by MaxAOC

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

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

Reply 5

mqb2766

Original post

by eugenemcmoy

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.

Reply 6

eugenemcmoy

Original post

by mqb2766

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 $e^{x^2+3x}$
$\dfrac{dv}{dx} = e^{x^2+3x}$ $/frac{1}{(2x+3)} e^{x^2+3x}$

Is the bottom line of working correct? I used the general rule $\int e^{ax+b} dx = \dfrac{1}{(a)} e^{ax+b}$

Reply 7

mqb2766

Original post

by eugenemcmoy

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 integrand does have two parts (multiplied together) and you have two integration techniques you could use

•

integration by parts which is based on the multiplication rule for differentiation (so multiplication in the original form)

•

integration by substitution / reverse chain rule which is based on the chain rule for differentation (so composition of the original form, rather than multiplication)

Looks like youve arbitrarily assumed its the former, as the second reply pretty much says and youve not considered the hint in the first reply. Almost by chance the last line in your working is really the way to think about the original integration problem.

Reply 8

DFranklin

Original post

by eugenemcmoy

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.

Reply 9

eugenemcmoy

Original post

by DFranklin

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 $e^{x^2}$

$\int e^{x^2+3x} dx$

Reply 10

mqb2766

Original post

by eugenemcmoy

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

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.

Reply 11

davros

Original post

by eugenemcmoy

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

To answer your question very specifically, you can't integrate $e^{x^2}$ - that should be an indicator that you shouldn't be trying!

You must have covered integration by substitution, so if you can't integrate $e^{x^2}$ but you do know how to integrate $e^{u}$ (with respect to u), what can you possibly do that turns $e^{x^2}$ into $e^{u}$ ?

Although integration in general is very difficult (and sometimes impossible), there aren't really many ways you can go wrong at A level, because there are only a few techniques available to you - recognition, IBP and substitution :smile:

Reply 12

DFranklin

Original post

by eugenemcmoy

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

As Davros said, you don't.

Nor, for that matter, do you want to integrate:

$\int e^{x^2+3x} dx$

Your actual integral is $\int (4x+6) e^{x^2+3x}\,dx$

I previously posted: if $\displaystyle \int f(x)\,dx = F(x)$ , then $\displaystyle \int g'(x) f(g(x)) \,dx = F(g(x))$ (**).

Note in particular, the f(g(x)) in (**). In other words, a function of a function. What, in your integral, looks a bit like a function of a function? So what might f be? And what might g be?

And then what is g'(x) f(g(x))? And does this have any relationship with the thing you're trying to integrate?

Reply 13

eugenemcmoy

Original post

by mqb2766

Yes I notice that $4x+6$ could be the derivative of
$e^{x^2+3x}$

Can you proceed with the working.

Reply 14

mqb2766

Original post

by eugenemcmoy

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

If you follow the hint in pretty much all the previous replies, its not integration by parts (which you tried to do a few times) but a simple substitution or equivalently the reverse chain rule for an exponential which you can pretty much write down.

If you really cant see it, think what would be the derivative of e^(x^2), then a simple transformation to the actual question.

(edited 1 year ago)

Reply 15

davros

Original post

by eugenemcmoy

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

Not being funny, but as per the rules of the forum, you should be doing the working - various people have provided you with hints to get you started, but you seem unwilling to have a go.

I'll be a bit more direct - at A level you've basically got 3 options when confronted with an integral:
(1) Do I recognise it? e.g. I know the derivative of sin x is cos x, so the integral of cos x better be sin x (+ constant);
(2) Does integration by parts look like a a productive strategy? In other words, can I express the integrand as uv where one of u, v is going to be differentiated, and the other function is going to be integrated, and crucially - will this make the integral simpler or more complicated? In your case, you can't integrate the exponential because it's not in a form you know how to integrate, but if you differentiate the exponential you'll bring down another (polynomial) multiplier and you'll also need to integrate the original polynomial in front of the exponential and multiply by that as well, so things are going to get much more complicated each time you apply IBP;
(3) can I find a substitution u = f(x) which will make the most complicated part of the integral look simpler, without causing an awful mess when I replace dx with (dx/du)du. Now, in your case, you don't know how to integrate e^(x^2 + 3x) dx, but you do know how to integrate e^u du. So what's the "obvious" thing to try?

Note that the so-called "reverse chain rule" is nothing more than a combination of substitution and/or recognition rolled into one step, but people make such a mess of it or mis-apply it that it really shouldn't be in your mind until you've done an awful lot of standard integrals using methods (1) - (3) above and can see what's going on :smile: