Integration by partial fractions with limits

[anthonyk2013]

\int (x+1/x²-3x-5)dx

I can't put the limits on the integral sign, 5 is the top limit and 3 is the bottom limit.

I can solve using partial fractions ok but I have never solved with limits before.

Where do the limits come in, do I need them at the start or can I factorise as usual and use the limits later in the solution?

 

[Mark44]

\int (x+1/x²-3x-5)dx

I can't put the limits on the integral sign, 5 is the top limit and 3 is the bottom limit.

I can solve using partial fractions ok but I have never solved with limits before.

Where do the limits come in, do I need them at the start or can I factorise as usual and use the limits later in the solution?

You have used parentheses, but they're not where they need to be. When a numerator or denominator consists of multiple terms, put parentheses around the entire numerator and/or the entire denominator. I gather that the integrand is (x + 1)/(x² - 3x - 5). As you wrote the integrand, it would be the same as x + (1/x²) - 3x - 5, which isn't what you intended.

To write the integral with limits in LaTeX, you can do it like this:

$$ \int_3^5 \frac{x + 1}{x^2 - 3x - 5}dx$$

I prefer to use $$ instead of [ tex ]. For inline LaTeX, you can use ## instead of [ itex ].

To answer your question, all you're doing when you use partial fractions is rewriting the integrand in a more convenient form. You bring the limits of integration along until you actually get the antiderivative.

For this problem, I probably wouldn't use partial fractions, since the denominator isn't going to factor into nice linear factors with integer coefficients. I'm not saying it won't work, but it will be a little messy.

 

[anthonyk2013]

You have used parentheses, but they're not where they need to be. When a numerator or denominator consists of multiple terms, put parentheses around the entire numerator and/or the entire denominator. I gather that the integrand is (x + 1)/(x² - 3x - 5). As you wrote the integrand, it would be the same as x + (1/x²) - 3x - 5, which isn't what you intended.

To write the integral with limits in LaTeX, you can do it like this:

$$ \int_3^5 \frac{x + 1}{x^2 - 3x - 5}dx$$

I prefer to use $$ instead of [ tex ]. For inline LaTeX, you can use ## instead of [ itex ].

To answer your question, all you're doing when you use partial fractions is rewriting the integrand in a more convenient form. You bring the limits of integration along until you actually get the antiderivative.

For this problem, I probably wouldn't use partial fractions, since the denominator isn't going to factor into nice linear factors with integer coefficients. I'm not saying it won't work, but it will be a little messy.

Thanks