# Integration question

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• December 20th 2009, 09:25 PM
Paymemoney
Integration question
Hi
Can someone tell me how would i anti-differentiate the following:
1) $\frac{x}{x+1}$
2) In this question you have to evaluate the following definite integrals:
This is what i done:
$\int_1^5e+\frac{1}{x}dx$
$=e^x+ln(x)$
$=e^5+ln^5-e^1+ln(1)$
$=ln(5)+e^4$

answer was ln(5) +4e but i don't understand how they got it.

P.S
• December 20th 2009, 09:42 PM
Chris L T521
Quote:

Originally Posted by Paymemoney
Hi
Can someone tell me how would i anti-differentiate the following:
1) $\frac{x}{x+1}$

Hint: $\frac{x}{x+1}=1-\frac{1}{x+1}$

Quote:

2) In this question you have to evaluate the following definite integrals:
This is what i done:
$\int_1^5e+\frac{1}{x}dx$
$=e^x+ln(x)$
$=e^5+ln^5-e^1+ln(1)$
$=ln(5)+e^4$

answer was ln(5) +4e but i don't understand how they got it.

P.S
$\int e\,dx\neq e^x$. Here, $e$ is a constant; this means we apply the rule $\int a\,dx=ax+C$.

Also, $e^5-e^1\neq e^4$. You can't subtract the exponents like that.
• December 20th 2009, 09:51 PM
Paymemoney
ok i understand why it is 4e now thanks.
• December 20th 2009, 11:45 PM
Paymemoney
Just be make sure i am right, for 1) do you divide x by x+1? or divide x+1 by x?
• December 21st 2009, 09:32 AM
oblixps
Quote:

Originally Posted by Paymemoney
Just be make sure i am right, for 1) do you divide x by x+1? or divide x+1 by x?

you divide x by x+1. use polynomial long division. or since these are simple and easy polynomials, you can use a little trick. what i do is add and subtract a 1 from the top so you have (x + 1 - 1) / (x + 1), then i rearrange: (x+1)/(x+1) - (1/(x+1)) so now the integrand becomes 1 - (1/(x+1))