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by **mitnord** on Wed Mar 25, 2009 9:20 pm

Please, help me solve this intergral:

Int (x^2 + 2x - 3) / (x + 1) dx,

when x > -1

Thank you.

**Sponsor**

by **stapel_eliz** on Wed Mar 25, 2009 9:38 pm

mitnord wrote:Int (x^2 + 2x - 3) / (x + 1) dx, when x > -1

By doing the long division, the integrand becomes:

. . . . . $x\, +\, 1\, +\, \frac{-4}{x\, +\, 1}$

The polynomial part is simple enough to integrate. :wink:

For the rational part, a simple substitution should work:

. . . . . $u\, =\, x\, +\, 1\mbox{, so }\, du\, =\, dx$

Then integrate (-4)(1/u), and back-substitute.

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How to solve this integral? int (x^2 + 2x - 3) / (x + 1) dx

How to solve this integral? int (x^2 + 2x - 3) / (x + 1) dx