Can someone please show and explain step by step how to evaluate this integral?
Problem #1)∫ 1/(x^3+x^2+x+1) dx
This is what I got so far. I am not sure what to do thereafter after doing partial fraction decomposition (in which I got A = 1/2, B = 1/2, and C = 1/2)
1/(x^2+1)(x+1) = ∫(-1/2x)+(1/2)/(x^2+1) + ((1/2)/(x+1)) dx
First step:
decompose into partial fractions (probably the hardest part of the problem)
Factorize the denominator (if possible)
x^3+x^2+x+1=x^2(x+1)+1(x+1)=(x^2+1)(x+1)
so we can decompose the given fraction into:
1/(x^3+x^2+x+1)
=1/2(x+1) + (1-x)/2(x^2+1)
which can be integrated readily.
To evaluate the integral ∫ 1/(x^3+x^2+x+1) dx, you have already correctly performed the partial fraction decomposition. You obtained A = 1/2, B = 1/2, and C = 1/2.
The next step is to rewrite the integral as a sum of simpler integrals:
∫ 1/(x^3+x^2+x+1) dx = ∫(-1/2x + 1/2)/(x^2+1) dx + ∫(1/2)/(x+1) dx
Let's solve each integral separately:
For the first integral, ∫(-1/2x + 1/2)/(x^2+1) dx, you can break it into two separate integrals:
∫(-1/2x)/(x^2+1) dx + ∫(1/2)/(x^2+1) dx
Now we can solve each integral separately:
For ∫(-1/2x)/(x^2+1) dx:
To integrate this type of integral, let's use the substitution method.
Let u = x^2 + 1. Differentiating both sides with respect to x, we get du = 2x dx. Rearranging, we have dx = du/(2x).
Substituting these into the integral, we have:
∫(-1/2x)/(x^2+1) dx = ∫(-1/2) * (1/u) * (du/(2x))
= -1/4 ∫du/u
Integrating the last expression gives:
= -1/4 ln|u| + C
Substituting u = x^2 + 1 back in:
= -1/4 ln|x^2 + 1| + C1
For the second integral, ∫(1/2)/(x^2+1) dx:
This is a standard integral. We recognize it as the arctangent function.
∫(1/2)/(x^2+1) dx = (1/2)∫dx/(x^2+1)
= (1/2) arctan(x) + C2
Now let's move on to the second part of the integral:
∫(1/2)/(x+1) dx = (1/2) ln|x+1| + C3
To obtain the final solution, sum up all the integrals:
∫ 1/(x^3+x^2+x+1) dx = -1/4 ln|x^2 + 1| + 1/2 arctan(x) + (1/2) ln|x+1| + C
And that's how you evaluate the given integral step by step.