# intergration with partial fractions

• July 31st 2007, 02:29 PM
davecs77
intergration with partial fractions
I am doing this problem:
integral of (x^3 + x - 1)/((X^2 +1))^2 dx
So i set that integral equal to (Ax + B)/(x^2 + 1) + (Cx + D)/((x^2 + 1))^2
I got ax^3 + ax + bx^2 + b + cx + d = 1x^3 + 1x - 1
so..
x^3 terms: a = 1
x^2 terms: b = 0
x^1 terms: a + c = 1 ( c = 0)
x^0 terms: b + d = -1 ( d = -1)
I put these back into the equation and got
integral of 1/(x^2 + 1) + integral of -1/((x^2 + 1))^2
which equals ln(x^2 + 1) + 1/(x^2 + 1) + C
The answer in the book has those 2 first terms as the answer..but has 2 more terms, then the Constant..I am having trouble finding the correct answer. What did I do wrong?
• July 31st 2007, 02:34 PM
Krizalid
First

$\int {\frac{{x^3 + x - 1}}
{{\left( {x^2 + 1} \right)^2 }}~dx} = \int {\frac{{x\left( {x^2 + 1} \right) - 1}}
{{\left( {x^2 + 1} \right)^2 }}~dx} = \int {\frac{x}
{{\left( {x^2 + 1} \right)}}~dx} - \int {\frac{1}
{{\left( {x^2 + 1} \right)^2 }}~dx}$

The second integral is already a partial fraction, apply integration by parts.
• July 31st 2007, 02:38 PM
davecs77
Quote:

Originally Posted by Krizalid
First

$\int {\frac{{x^3 + x - 1}}
{{\left( {x^2 + 1} \right)^2 }}~dx} = \int {\frac{{x\left( {x^2 + 1} \right) - 1}}
{{\left( {x^2 + 1} \right)^2 }}~dx} = \int {\frac{x}
{{\left( {x^2 + 1} \right)}}~dx} - \int {\frac{1}
{{\left( {x^2 + 1} \right)^2 }}~dx}$

The second integral is already a partial fraction, apply integration by parts.

wait..what did i do wrong? I am confused on what you are telling me.
• July 31st 2007, 03:05 PM
topsquark
Quote:

Originally Posted by davecs77
I am doing this problem:
integral of (x^3 + x - 1)/((X^2 +1))^2 dx
So i set that integral equal to (Ax + B)/(x^2 + 1) + (Cx + D)/((x^2 + 1))^2
I got ax^3 + ax + bx^2 + b + cx + d = 1x^3 + 1x - 1
so..
x^3 terms: a = 1
x^2 terms: b = 0
x^1 terms: a + c = 1 ( c = 0)
x^0 terms: b + d = -1 ( d = -1)
I put these back into the equation and got
integral of 1/(x^2 + 1) + integral of -1/((x^2 + 1))^2
which equals ln(x^2 + 1) + 1/(x^2 + 1) + C
The answer in the book has those 2 first terms as the answer..but has 2 more terms, then the Constant..I am having trouble finding the correct answer. What did I do wrong?

Quote:

Originally Posted by Krizalid
First

$\int {\frac{{x^3 + x - 1}}
{{\left( {x^2 + 1} \right)^2 }}~dx} = \int {\frac{{x\left( {x^2 + 1} \right) - 1}}
{{\left( {x^2 + 1} \right)^2 }}~dx} = \int {\frac{x}
{{\left( {x^2 + 1} \right)}}~dx} - \int {\frac{1}
{{\left( {x^2 + 1} \right)^2 }}~dx}$

The second integral is already a partial fraction, apply integration by parts.

Quote:

Originally Posted by davecs77
wait..what did i do wrong? I am confused on what you are telling me.

Both you are Krizalid are coming up with the same integrals. So look to the second integral for possible problems. For reference, my calculator gives:
$\frac{ln(x^2 + 1)}{2} - \frac{1}{2}tan^{-1}(x) - \frac{x}{2(x^2 + 1)} + C$

-Dan