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

Answer 1

#intx^3/(x^2+2x+1)dx=x^2-2x+3ln|x+1|+1/(x+1)#

#x^3/(x^2+2x+1)=(x(x^2+2x+1)-2(x^2+2x+1)+3x+2)/(x^2+2x+1)#
= #x-2+(3x+2)/(x^2+2x+1)#. Let
#(3x+2)/(x^2+2x+1)=(3x+2)/(x+1)^2-=A/(x+1)+B/(x+1)^2#
i.e. #3x+2=A(x+1)+B#
Now if #x=0#, we have #A+B=2# and if #x=-1#, we have #B=-1# and hence #A=3#
Therefore #(3x+2)/(x^2+2x+1)=(3x+2)/(x+1)^2-=3/(x+1)-1/(x+1)^2#
and #x^3/(x^2+2x+1)=x-2+3/(x+1)-1/(x+1)^2#
hence #intx^3/(x^2+2x+1)dx=int(x-2+3/(x+1)-1/(x+1)^2)dx#
= #x^2-2x+3ln|x+1|+1/(x+1)#
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Answer 2

# (x+1)^2/2-3(x+1)+3ln|x+1|+1/(x+1)+C.#

Though the Problem is requred to be solved using Partial Fractions,

as a Second Method, we show that the same can be solved

easily without its use.

Let, #I=intx^3/(x^2+2x+1)dx=intx^3/(x+1)^2dx.#
Sbst.ing, #x+1=t," so that, "dx=dt, and, x=t-1.#
#:. I=int(t-1)^3/t^2dt,#
#=int{t^3-1-3t(t-1)}/t^2dt,#
#=int{t^3-3t^2+3t-1}/t^2dt,#
#=int{t^3/t^2-3t^2/t^2+3t/t^2-1/t^2}dt,#
#=int{t-3+3/t-1/t^2}dt,#
#=t^2/2-3t+3ln|t|+1/t,#
# rArr I=(x+1)^2/2-3(x+1)+3ln|x+1|+1/(x+1)+C.#

Enjoy Maths.!

N.B.: #I=x^2/2+x+1/2-3x-3+3ln|x+1|+1/(x+1)+C,#
#=x^2/2-2x-5/2+3ln|x+1|+1/(x+1)+C,#
#=x^2/2-2x+3ln|x+1|+1/(x+1)+c, where, c=C-5/2,# as

Respected Shwetank Mauria has derived!

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Answer 3

To integrate ( \frac{x^3}{x^2 + 2x + 1} ) using partial fractions, first factor the denominator, which is ( (x + 1)^2 ). Then, express the fraction as ( \frac{A}{x+1} + \frac{B}{(x+1)^2} ). After finding the values of ( A ) and ( B ), integrate each term separately. The integral of ( \frac{A}{x+1} ) can be found using the natural logarithm, and the integral of ( \frac{B}{(x+1)^2} ) can be found using reverse power rule.

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Answer from HIX Tutor

When evaluating a one-sided limit, you need to be careful when a quantity is approaching zero since its sign is different depending on which way it is approaching zero from. Let us look at some examples.

When evaluating a one-sided limit, you need to be careful when a quantity is approaching zero since its sign is different depending on which way it is approaching zero from. Let us look at some examples.

When evaluating a one-sided limit, you need to be careful when a quantity is approaching zero since its sign is different depending on which way it is approaching zero from. Let us look at some examples.

When evaluating a one-sided limit, you need to be careful when a quantity is approaching zero since its sign is different depending on which way it is approaching zero from. Let us look at some examples.

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