Question

Find the following indefinite integral : integral (1+x^2)sin(x^3+3x)dx

Answer 1

\[\int\limits(1+x^2)\sin(x^3+3x)dx\]

Answer 2

Hey man, I'm a biomed engineer, just revising some integration.

Listen, I know the basics here, but when I got stuck in the question,

can I ask you about where I got stuck?

Answer 3

Heres how far i got:

Let u = x^3 + 3x

du/dx = 3x^2+3 => dx = du/3x^2+3

From here?

Answer 4

Then just replace dx in your integral with du/3x² + 3
Better yet, take notice of the fact that a constant can be factored out of 3x² + 3

Answer 5

I got \[\int\limits \sin(u) . (1+x^2) . du/3x^2+3\]

Answer 6

Hang on, easier to see it like this:
\[\int\limits\limits(1+x^2)\sin(x^3+3x)\frac{du}{3 + 3x^{2}}\]

Answer 7

So, the denominator of 3 + 3x², 3 can be factored out, right?

Answer 8

1/3 ?

Answer 9

Outside the integral

Answer 10

That's right ;
\[\int\limits\limits\limits\sin(x^3+3x)\frac{(1 + x^{2})du}{3(1 + x^{2})}\]
Consequently, as you said:
\[\frac{1}{3}\int\limits_{}^{} \sin(x^{3}+3x)du\]
Carry on? :D

Answer 11

So \[\frac{ 1 }{ 3 }\int\limits \sin(u) . du ?\]

Answer 12

Precisely :D

Answer 13

See, I'm excellent at method, its SEEING manipulations I'm weak at.

Comes with practice?

Answer 14

Indeed it does :D

Answer 15

Ok, do you mind if I ask you one other quick thing? Then I'm sorted.

Answer 16

Go ahead

Answer 17

\[\int\limits \frac{ 7dx }{ (x-4)(x+2) }\]

Answer 18

Partial fraction?

Answer 19

It would certainly seem like that's the way to go. You know where to begin?

Answer 20

I'm thinking bring the 7 outside the integral first?

Answer 21

You could do that:
\[7\int\limits_{}^{}\frac{dx}{(x-4)(x+2)}\]
And then?

Answer 22

Do partial fractions of 1/(x-4) + 1/(x+2)

Answer 23

Is there another method?

You can't do a substitution, and LIATE doesnt work?

Answer 24

Substitution doesn't seem possible, and I don't know what LIATE is o.O
And I'm afraid it's no simple matter of separating the fraction with 1 as the numerator on both sides.

Don't worry though, it may not be THAT simple, but it IS rather simple :)
Your factors (x + 2) and (x - 4) are both linear, or or degree 1. Then all we know about their respective numerators is that they'll be of one LESS degree, and therefore their numerators are just constants. Let's call them A and B.

\[\frac{1}{(x-4)(x+2)}=\frac{A}{x-4}+\frac{B}{x+2}\]
Can you do it from here?

Answer 25

Yes, I can do that. I'm just curious to how you will then integrate?

By the way, LIATE = Logs, inverse trig, algebra, trig, exponential,

When looking at a question, I see if it has any of these, to decide what to do.

Answer 26

Well, integration distributes over a sum, and integrating a constant over a linear function is simple, to illustrate

Answer 27

\[\int\limits_{}^{}\frac{k}{ax+b}dx=\frac{k}{a}\ln|ax + b| +C\]

Answer 28

Yeah, was going to mention log would come into it.

Will we end up with a LogA - Log B?

Answer 29

No. Let's carry on from \[\frac{1}{(x-4)(x+2)}=\frac{A}{x-4}+\frac{B}{x+2}\]
\[=\frac{A(x+2)+B(x-4)}{(x-4)(x+2)}\]

Answer 30

Ok

Answer 31

\[=\frac{(A+B)x+2A-4B}{(x-4)(x+2)}=\frac{1}{(x-4)(x+2)}\]
Is it clearer from here?

Answer 32

Yes

Answer 33

Well, can you do it from there?

Answer 34

Pretty sure i can! :D

Answer 35

Thanks, mate :)

Answer 36

No problem