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

Question

Elijah Abram · Accepted Answer

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

We split the integrand using partial fractions before integrating the function.

#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)#

Integrating now, we obtain

#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"#

Ezekiel Alexander · Answer

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

Dividends in Parts

#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)#

enlarge the brackets

#3=Ax+A+Bx-2B#

factorize

#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)#

Combination

#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#

owing to the ongoing multiple of -1

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

Separately assess the two integrals.

#int1/(x-2)dx#

Apply the u-substitution method.

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

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

Introduce the integral to the world of u

#int1/udu#

#ln|\u|+C_1#

Return to the world of X

#ln|\x-2|+C_1#

the second integral

#-int1/(x+1)dx#

To avoid confusion for both you and the marker, substitute a different letter.

#v=x+1#

#(dv)/dx=1#

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

introduce the integral to the world of v

#-int1/vdv#

#-(ln|\v|+C_2)#

return to the world of X

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

We will simply create another constant since a negative constant, added or subtracted from another constant, is still a 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#

reunite the components of the integral that we addressed.

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

#:.#

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

Elijah Abram · Answer

undefined To integrate $ \frac{3}{(x-2)(x+1)} $ using partial fractions, first express it as a sum of partial fractions with undetermined coefficients:

$$ \frac{3}{(x-2)(x+1)} = \frac{A}{x-2} + \frac{B}{x+1} $$

Multiply both sides by $ (x-2)(x+1) $ to clear the denominators:

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

Expand and group like terms:

$$ 3 = Ax + A + Bx - 2B $$

Now, equate coefficients of like terms:

For x terms: $ A + B = 0 $  
For constant terms: $ A - 2B = 3 $

Solve these equations simultaneously to find the values of A and B.

From the first equation, $ A = -B $, substitute into the second equation:

$$ -B - 2B = 3 $$
$$ -3B = 3 $$
$$ B = -1 $$

Then, substitute $ B = -1 $ into $ A = -B $:

$$ A = -(-1) = 1 $$

So, the partial fraction decomposition is:

$$ \frac{3}{(x-2)(x+1)} = \frac{1}{x-2} - \frac{1}{x+1} $$

Now, integrate each term separately:

$$ \int \frac{3}{(x-2)(x+1)} \, dx = \int \frac{1}{x-2} \, dx - \int \frac{1}{x+1} \, dx $$

$$ = \ln|x-2| - \ln|x+1| + C $$

$$ = \ln\left|\frac{x-2}{x+1}\right| + C $$