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

Question

Isabella Ackerman · Accepted Answer

The answer is #=-1/(2(x+1)^2)+(2)/(x+1)+ln(x+1)+C# Let's do the decomposition in partial fractions #x^2/(x+1)^3=A/(x+1)^3+B/(x+1)^2+C/(x+1)# #=(A+B(x+1)+C(x+1)^2)/(x+1)^3# We need to find A,B,C So #x^2=A+B(x+1)+C(x+1)^2# coefficients of #x^2##=>##C=1# let #x=-1##=>##A=1# Coefficients of x, #0=B+2C##=>##B=-2# #:.x^2/(x+1)^3=1/(x+1)^3-2/(x+1)^2+1/(x+1) # So #int(x^2dx)/(x+1)^3=intdx/(x+1)^3-int(2dx)/(x+1)^2+intdx/(x+1)# We use #intx^ndx=x^(n+1)/(n+1)# and #intdx/x=lnx# #int(x^2dx)/(x+1)^3=-1/(2(x+1)^2)+(2)/(x+1)+ln(x+1)+C#

Christopher Allers · Answer

#int color(white)(.)x^2/(x+1)^3color(white)(.) dx = ln abs(x+1)+2/(x+1)-1/(2(x+1)^2)+C# #int color(white)(.)x^2/(x+1)^3color(white)(.) dx = int color(white)(.)((x^2+2x+1)-(2x+2)+1)/(x+1)^3color(white)(.) dx# #color(white)(int color(white)(.)x^2/(x+1)^3color(white)(.) dx) = int color(white)(.)((x+1)^2-2(x+1)+1)/(x+1)^3color(white)(.) dx# #color(white)(int color(white)(.)x^2/(x+1)^3color(white)(.) dx) = int color(white)(.)1/(x+1)-2/(x+1)^2+1/(x+1)^3color(white)(.) dx# #color(white)(int color(white)(.)x^2/(x+1)^3color(white)(.) dx) = ln abs(x+1)+2/(x+1)-1/(2(x+1)^2)+C#

Samantha Adkison · Answer

undefined To integrate $ \frac{x^2}{(x+1)^3} $ using partial fractions, first express the rational function as a sum of simpler fractions with undetermined coefficients. In this case, we write:

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

Then, clear the denominators by multiplying both sides by $ (x+1)^3 $ to obtain:

$$ x^2 = A(x+1)^2 + B(x+1) + C $$

Now, you can solve for $ A $, $ B $, and $ C $ by expanding and matching coefficients. Once you find the values of $ A $, $ B $, and $ C $, you can integrate each term separately. Finally, sum up the integrals to find the result.