Help me integrate this please :)

Original post by alpha_4

Note that

\dfrac{(x+1)^2}{x^2+1} \equiv \dfrac{(x^2+1) + 2x}{x^2+1}

and then recall that

\dfrac{a+b}{c} \equiv \dfrac{a}{c} + \dfrac{b}{c}

. This will leave you with two trivial integrals to deal with.

(edited 13 years ago)

Reply 2

Sockpirate

Try using long division first and see where that gets you :smile:

EDIT: Sorry, why was this negged? What I said achieves exactly the same as what Farhan said - divide the top bit by the bottom bit (hence long division), to get an expression that's much simpler to integrate. I was just trying to help :facepalm:

(edited 13 years ago)

Reply 3

spex

\frac{(x+1)^2}{x^2+1} = \frac{x^2 +1 +2x}{x^2+1}

Reply 4

Original post by Farhan.Hanif93

Note that

\dfrac{(x+1)^2}{x^2+1} \equiv \dfrac{(x^2+1) + 2x}{x^2+1}

and then recall that

\dfrac{a+b}{c} \equiv \dfrac{a}{c} + \dfrac{b}{c}

. This will leave you with two trivial integrals to deal with.

thank you very much

Reply 5

Original post by Farhan.Hanif93

Note that

\dfrac{(x+1)^2}{x^2+1} \equiv \dfrac{(x^2+1) + 2x}{x^2+1}

and then recall that

\dfrac{a+b}{c} \equiv \dfrac{a}{c} + \dfrac{b}{c}

. This will leave you with two trivial integrals to deal with.

Sorry to bother you, but what would you do for this integral:
1/(x^2)(x-1)

thanks :biggrin:

Reply 6

Original post by alpha_4

Sorry to bother you, but what would you do for this integral:
1/(x^2)(x-1)

thanks :biggrin:

Use partial fractions.

Reply 7

Original post by alpha_4

Sorry to bother you, but what would you do for this integral:
1/(x^2)(x-1)

thanks :biggrin:

Partial fractions is probably a good idea.

Express

\dfrac{1}{x^2 (x-1)}

in the form

\dfrac{Ax+B}{x^2} + \dfrac{C}{x-1}

, and find the constants A, B and C.

Reply 8

Original post by Farhan.Hanif93

Partial fractions is probably a good idea.

Express

\dfrac{1}{x^2 (x-1)}

in the form

\dfrac{Ax+B}{x^2} + \dfrac{C}{x-1}

, and find the constants A, B and C.

How about this ?

\frac 1{x^2(x-1)}=\frac {\large A}{x^2} + \frac {\large B}{x}+ \frac {\large C}{x-1}

Oh wait. same thing. Sorry

Reply 9

Original post by CookieGhoul

Oh wait. same thing. Sorry

Yep.

Reply 10

Original post by Farhan.Hanif93

Partial fractions is probably a good idea.

Express

\dfrac{1}{x^2 (x-1)}

in the form

\dfrac{Ax+B}{x^2} + \dfrac{C}{x-1}

, and find the constants A, B and C.

Thanks, I don't think those types of partial fractions are in the c4 syllabus but it seems like the most sensible option :smile:

Only one more, if you don't mind...

x^2/X^3 + 1

Reply 11

Original post by alpha_4

Thanks, I don't think those types of partial fractions are in the c4 syllabus but it seems like the most sensible option :smile:

Only one more, if you don't mind...

x^2/X^3 + 1

Is it (x^2)/(x^3+1) or (x^2)/(x^3) + 1

Reply 12

Original post by alpha_4

Thanks, I don't think those types of partial fractions are in the c4 syllabus but it seems like the most sensible option :smile:

Only one more, if you don't mind...

x^2/X^3 + 1

Note that

\dfrac{x^2}{x^3+1} \equiv \dfrac{1}{3} \times \dfrac{3x^2}{x^3+1} \equiv \dfrac{1}{3} \times \dfrac{\frac{d}{dx}[x^3+C]}{x^3+1}

. What do you know about integrals of that form. If that method doesn't work for you, use a substitution of

u=x^3-1

Original post by CookieGhoul

Is it (x^2)/(x^3+1) or (x^2)/(x^3) + 1

I'd assume that he meant the former as the latter is reasonably trivial.

(edited 13 years ago)

Reply 13

Original post by CookieGhoul

Is it (x^2)/(x^3+1) or (x^2)/(x^3) + 1

(x^2)/(x^3) + 1
the latter, sorry!

Reply 14

Original post by Farhan.Hanif93

Note that

\dfrac{x^2}{x^3+1} \equiv \dfrac{1}{3} \times \dfrac{3x^2}{x^3+1} \equiv \dfrac{1}{3} \times \dfrac{\frac{d}{dx}[x^3+C]}{x^3+1}

. What do you know about integrals of that form. If that method doesn't work for you, use a substitution of

u=x^3-1

I'd assume that he meant the former as the latter is reasonably trivial.

Didn't you just integrate the former?

Reply 15

Original post by alpha_4

(x^2)/(x^3) + 1
the latter, sorry!

Omg dude. Then you'll be integrating

\dfrac {1}{x} + 1

Reply 16

Original post by alpha_4

(x^2)/(x^3) + 1
the latter, sorry!

You need to be using LaTeX to make sure your posts are clear. If you can't do that integral, you need to reopen your textbook and learn about integrals that lead to logs properly. Then come back and ask for help if you're stuck.

Reply 17

Original post by Farhan.Hanif93

LOL! + rep to you :biggrin:

Haha, you wrote so much for the other case, and he just wanted a simple integral

Reply 18