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## #1 2005-10-14 01:40:07

Student
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### Integration by substitution

Can somebody please help me!
∫ x^2/1 - 2x^3

## #2 2005-10-14 07:39:37

MathsIsFun
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### Re: Integration by substitution

As a start, that should probably be:

∫ x^2/(1 - 2x^3) dx

is that right?

Otherwise it is too easy

"The physicists defer only to mathematicians, and the mathematicians defer only to God ..."  - Leon M. Lederman

## #3 2005-10-14 13:56:26

Flowers4Carlos
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### Re: Integration by substitution

if u think that is too easy try this one:

∫(x²-29x+5)/[(x-4)²(x²+3)]dx

this tuff question was on my calculus2 final, and no, i did not solve it

## #4 2005-10-14 15:42:45

MathsIsFun
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### Re: Integration by substitution

Ouch, that is a tough one!

But I didn't get around to solving "Student"s integral. Which I will attempt now ...

∫ x^2/(1 - 2x^3) dx

Try: u=1-2x^3
Then: du = -6x^2 dx (ie we differentiated it)
So: x^2 dx = (-1/6) du

Substituting 1-2x^3=u and x^2 dx = (-1/6) du we get:

∫ x^2/(1 - 2x^3) dx =  ∫ 1/u (-1/6) du

Then it is fairly easy to solve ∫ 1/u (-1/6) du:

∫ 1/u (-1/6) du = (-1/6) ∫ 1/u  du = (-1/6) ln u

Now put back u=1-2x^3:

(-1/6) ln( 1-2x^3 )

DONE! (I hope!)

"The physicists defer only to mathematicians, and the mathematicians defer only to God ..."  - Leon M. Lederman

## #5 2005-10-14 17:02:47

ganesh
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### Re: Integration by substitution

A little further.....
(-1/6) ln( 1-2x^3 )=
ln[1/(1-2x^3)^1/6]
MathsIsFun had done it right, I posted this
.... just to complicate things

Character is who you are when no one is looking.

## #6 2005-10-14 17:20:25

Flowers4Carlos
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### Re: Integration by substitution

... to complicate it even further:
ln[1/(1-2x^3)^1/6] + C

## #7 2005-10-14 19:11:12

MathsIsFun
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### Re: Integration by substitution

Yes, correct! The constant is necessary for completeness.

∫ x^2/(1 - 2x^3) dx = (-1/6) ln( 1-2x^3 ) + C

"The physicists defer only to mathematicians, and the mathematicians defer only to God ..."  - Leon M. Lederman

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