How do you integrate 3/((x-2)(x+1)) using partial fractions?

2 Answers
Mar 5, 2018

int3/((x-2)(x+1))dx=lnabs((x-2)/(x+1))+"c"

Explanation:

In order to integrate the function, we first use partial fractions to split the integrand

3/((x-2)(x+1))=3 1/((x-2)(x+1))=3 ((x+1)-(x-2))/(3(x-2)(x+1))=3 (((x+1))/(3(x-2)(x+1))-((x-2))/(3(x-2)(x+1)))=1/(x-2)-1/(x+1)

Now integrating, we get

3/((x-2)(x+1))dx=int1/(x-2)-1/(x+1)dx=lnabs(x-2)-lnabs(x+1)+"c"=lnabs((x-2)/(x+1))+"c"

Mar 6, 2018

int3/((x-2)(x+1))=ln|\x-2|+ln|\1/(x+1)|+C

Explanation:

Partial Fractions

3/((x-2)(x+1))=A/(x-2)+B/(x+1)

Multiply the whole equation by (x-2)(x+1)

3=A(x+1)+B(x-2)

expand the brackets

3=Ax+A+Bx-2B

factorise

3=x(A+B)+A-2B

A+B has to equal 0 because there is no x term for the other side
and A-2B has to equal 3 because those terms have no x same as 3

So

A+B=0 (1)
and
A-2B=3 (2)

subtract equation (2) from equation (1) to get

0-3B=3
(cancel(-3)B)/cancel(-3)=-cancel((3)/3)

:. B=-1

Substitute B for 3 in either equation (1) or (2) to get A

A+(-1)=0
:.A=1

or

A-2(-1)=3
:.A=1

:. 3/((x-2)(x+1))=1/(x-2)+(-1)/(x+1)

Integration

int3/((x-2)(x+1))dx=int1/(x-2)+(-1)/(x+1)dx

int1/(x-2)+(-1)/(x+1)dx=int1/(x-2)dx+int(-1)/(x+1)dx

int1/(x-2)dx+int(-1)/(x+1)dx=int1/(x-2)dx-int1/(x+1)dx

because of the constant multiple of -1

int1/(x-2)dx-int1/(x+1)dx

Evaluate the two integrals separately

int1/(x-2)dx

use u-substitution

u=x-2
(du)/dx=1

(cancel(dx)du)/cancel(dx)=1*dx

Bring the integral into the u world

int1/udu

ln|\u|+C_1

Bring back into the x world

ln|\x-2|+C_1

second integral

-int1/(x+1)dx

use another letter substitution so you or the marker doesn't get confused

v=x+1

(dv)/dx=1

(cancel(dx)dv)/cancel(dx)=1*dx

bring the integral into the v world

-int1/vdv

-(ln|\v|+C_2)

bring back into the x world

-(ln|\x+1|+C_2)

A negative constant, adding or subtracting a constant from another constant is still a constant so we will just make another constant

-ln|\x+1|+C_3

Use the log law alog_b(c)=log_b(c^a)

ln|\(x+1)^-1|+C_3

ln|\1/(x+1)|+C_3

put the parts of the integral that we answered back together

ln|\x-2|+ln|\1/(x+1)|+C_1+C_3

:.

ln|\x-2|+ln|\1/(x+1)|+C