Intergrating Fractions?

[MickGrif]

Hi, I'm trying to understand how you integrate fractions, I have a couple of examples which I was given, unfortunately the lecturer skipped the actual method and just gave us the answers

.
By any chance could someone go through the method to solve these? (I think you use U substitution?)
Thanks in advance

(a) integral ((1-z)/(z² +1))dz

(b) integral ((1)/(z+z²)) dz

(c) integral ((z)/(1-3z²))dz

 

[HallsofIvy]

Hi, I'm trying to understand how you integrate fractions, I have a couple of examples which I was given, unfortunately the lecturer skipped the actual method and just gave us the answers

.
By any chance could someone go through the method to solve these? (I think you use U substitution?)
Thanks in advance

(a) integral ((1-z)/(z² +1))dz

$\displaystyle \int \frac{1- z}{z^2+ 1}dz= \int \frac{1}{z^2+ 1}dz- \int\frac{z}{z^2+ 1}dz$
For the first, look at the derivative of arctan(x) and for the second, use the substitution [itex]u= z^2+ 1[/itex].

(b) integral ((1)/(z+z²)) dz

$\displaystyle \frac{1}{z+ z^2}= \frac{1}{z(1+ z)}$. Use "partial fractions" to write $\displaystyle \frac{1}{z(1+ z)}= \frac{A}{z}+ \frac{B}{1+z}$. Solve for A and B so that is true for all z.

(c) integral ((z)/(1-3z²))dz

Use the substitution $\displaystyle u= 1- 3z^2$

 

[MickGrif]

$\displaystyle \int \frac{1- z}{z^2+ 1}dz= \int \frac{1}{z^2+ 1}dz- \int\frac{z}{z^2+ 1}dz$
For the first, look at the derivative of arctan(x) and for the second, use the substitution [itex]u= z^2+ 1[/itex].


$\displaystyle \frac{1}{z+ z^2}= \frac{1}{z(1+ z)}$. Use "partial fractions" to write $\displaystyle \frac{1}{z(1+ z)}= \frac{A}{z}+ \frac{B}{1+z}$. Solve for A and B so that is true for all z.


Use the substitution $\displaystyle u= 1- 3z^2$

Thanks for the answer, quick question, could you do a partial fraction method on the first equation or would that not work?

 

[Deleted member 4993]

Thanks for the answer, quick question, could you do a partial fraction method on the first equation or would that not work?

Depends on how you plan to factorize (z² + 1)...