long complicated integral.

**gammaman** · April 20th 2009, 12:00 PM

Does anyone know any shortcut to solve the following integral? As it stands the only way I know how to do it is to first use polynomial long division and then use integration with partial fractions. It is not necessarily difficult, it is just long and tedious, so I was wondering if there were any short cuts to lessen the chances of me making a careless error.

$\int\frac{2x^3+7x^2-15x+18}{2x^2+5x-3}$

**skeeter** · April 20th 2009, 01:33 PM

Originally Posted by gammaman

Does anyone know any shortcut to solve the following integral? As it stands the only way I know how to do it is to first use polynomial long division and then use integration with partial fractions. It is not necessarily difficult, it is just long and tedious, so I was wondering if there were any short cuts to lessen the chances of me making a careless error.

$\int\frac{2x^3+7x^2-15x+18}{2x^2+5x-3}$

do the long division and the partial fractions ... yes it's tedious, but you probably need the algebra practice anyway.

**Nacho** · April 20th 2009, 01:38 PM

$ \begin{gathered} I = \int {\frac{{2x^3 + 7x^2 - 15x + 18}} {{2x^2 + 5x - 3}}dx = \int {\frac{{\left( {x + 1} \right)\left( {2x^2 + 5x - 3} \right) - 7x + 21}} {{2x^2 + 5x - 3}}dx} } \hfill \\ \hfill \\ I = \int {\left( {x + 1} \right)dx - \underbrace {\int {\frac{{7x - 21}} {{2x^2 + 5x - 3}}dx} }_A} \hfill \\ \end{gathered} $

$ A = \int {\frac{{7x - 21}} {{2x^2 + 5x - 3}}dx} = \frac{7} {4}\int {\frac{{4x - 12 + (5 - 5)}} {{2x^2 + 5x - 3}}dx} = $ $ \frac{7} {4}\int {\frac{{4x - 12 + 5}} {{2x^2 + 5x - 3}}dx} - \frac{7} {4}\int {\frac{{17}} {{2x^2 + 5x - 3}}dx} $

$ A = \frac{7} {4}\int {\frac{{\left( {2x^2 + 5x - 3} \right)^\prime }} {{2x^2 + 5x - 3}}dx} - \frac{{7 \cdot 17}} {4}\int {\frac{1} {{\left( {2x + \frac{5} {2}} \right)^2 - \frac{{37}} {4}}}dx} $

Can you continue?
Anything that you didn´t understand, only post it

**gammaman** · April 20th 2009, 01:49 PM

Originally Posted by skeeter

do the long division and the partial fractions ... yes it's tedious, but you probably need the algebra practice anyway.

Your right I could really use the algebra practice. The only thing I am not solid on is how to make sure that I factor

2x^2+5x-3 correctly. I know that it is (2x-1)(x+3) but only because when you foil, you see that it is correct. Is there any short-cut to check that you factored correctly without having to foil?

**Moo** · April 20th 2009, 01:52 PM

Hello,

Originally Posted by Nacho

$ \begin{gathered} I = \int {\frac{{2x^3 + 7x^2 - 15x + 18}} {{2x^2 + 5x - 3}}dx = \int {\frac{{\left( {x + 1} \right)\left( {2x^2 + 5x - 3} \right) - {\color{red}7}x + 21}} {{2x^2 + 5x - 3}}dx} } \hfill \\ \end{gathered} $

I'm afraid it's -17x

**Nacho** · April 20th 2009, 05:24 PM

Originally Posted by gammaman

2x^2+5x-3 correctly. I know that it is (2x-1)(x+3) but only because when you foil, you see that it is correct. Is there any short-cut to check that you factored correctly without having to foil?

$ 2x^2 + 5x - 3 = \frac{{2\left( {2x^2 + 5x - 3} \right)}} {2} = \frac{{\left( {2x} \right)^2 + 5\left( {2x} \right) - 6}} {2} = \frac{{\left( {2x + 6} \right)\left( {2x - 1} \right)}} {2} $

Originally Posted by Moo

I'm afraid it's -17x

Is true, thanks... but the importante is the idea

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