Integral Convergence and Divergence II

[ardentmed]

This thread is only for question 5.

As for number 5 part a, after tediously expanding the partial fraction expression, I ended up getting c=1, d=0, b=1, and c=1, ultimately resulting in: ln(x) - (1/x^2) + c. I really don't think this looks right.

As for 5b, I obtained b=-1, c=-1, a=2, and 2lnx - (1/2)ln(x^2 +3) - (1/3)arctan(x/√3)Thanks in advance.

 

[evinda]

This thread is only for question 5.

As for number 5 part a, after tediously expanding the partial fraction expression, I ended up getting c=1, d=0, b=1, and c=1, ultimately resulting in: ln(x) - (1/x^2) + c. I really don't think this looks right.

As for 5b, I obtained b=-1, c=-1, a=2, and 2lnx - (1/2)ln(x^2 +3) - (1/3)arctan(x/√3)Thanks in advance.

For the part $a$ it is like that:$$\frac{1}{x^2(x-1)^2}=\frac{A}{x}+\frac{B}{x^2}+\frac{C}{x-1}+\frac{D}{(x-1)^2}=\frac{Ax(x-1)^2+B(x-1)^2+Cx^2(x-1)+Dx^2}{x^2(x-1)^2}=\frac{Ax(x^2-2x+1)+B(x^2-2x+1)+C(x^3-x^2)+Dx^2}{x^2(x-1)^2}=\frac{Ax^3-2Ax^2+Ax+Bx^2-2Bx+B+Cx^3-Cx^2+Dx^2}{x^2(x-1)^2}=\frac{(A+C)x^3+(-2A+B-C+D)x^2+(A-2B)x+B}{x^2(x-1)^2}$$

Therefore,it must be:

$$B=1 \\ A-2B=0 \Rightarrow A=2 \\ A+C=0 \Rightarrow C=-2 \\ -2A+B-C+D=0 \Rightarrow -4+1+2+D=0 \Rightarrow D=1 $$

So,we have:

$$\int \frac{1}{x^2(x-1)^2}dx= \int \frac{2}{x} dx+ \int \frac{1}{x^2}dx- \int \frac{2}{x-1}dx+ \int \frac{1}{(x-1)^2}dx \\ =2 \ln |x|-\frac{1}{x}-2 \ln |x-1|-\frac{1}{x-1}$$

 

[evinda]

For part $b$,you found $a,b,c$ correct,as it is like that:

$$\frac{x^2-x+6}{x^3+3x}=\frac{x^2-x+6}{x(x^2+3)}=\frac{A}{x}+\frac{Bx+C}{x^2+3}=\frac{A(x^2+3)+x(Bx+C)}{x(x^2+3)}=\frac{(A+B)x^2+Cx+3A}{x(x^2+3)}$$

Therefore,it must be:

$$3A=6 \Rightarrow A=2 \\ C=-1 \\ A+B=1 \Rightarrow B=-1$$

Therefore:

$$\int \frac{x^2-x+6}{x^3+3x} dx= \int \frac{2}{x} dx- \int \frac{x+1}{x^2+3} dx$$

You will conclude to this: $$\int \frac{x^2-x+6}{x^3+3x} dx=-\frac{1}{2} \ln (x^2+3)+2 \ln x-\frac{arc \tan (\frac{x}{ \sqrt{3}})}{ \sqrt{3}}+c$$

 

[Prove It]

For part $b$,you found $a,b,c$ correct,as it is like that:

$$\frac{x^2-x+6}{x^3+3x}=\frac{x^2-x+6}{x(x^2+3)}=\frac{A}{x}+\frac{Bx+C}{x^2+3}=\frac{A(x^2+3)+x(Bx+C)}{x(x^2+3)}=\frac{(A+B)x^2+Cx+3A}{x(x^2+3)}$$

Therefore,it must be:

$$3A=6 \Rightarrow A=2 \\ C=-1 \\ A+B=1 \Rightarrow B=-1$$

Therefore:

$$\int \frac{x^2-x+6}{x^3+3x} dx= \int \frac{2}{x} dx- \int \frac{x+1}{x^2+3} dx$$

You will conclude to this: $$\int \frac{x^2-x+6}{x^3+3x} dx=-\frac{1}{2} \ln (x^2+3)+2 \ln x-\frac{arc \tan (\frac{x}{ \sqrt{3}})}{ \sqrt{3}}+c$$

Just a LaTeX tip, \arctan produces $\displaystyle \begin{align*} \arctan \end{align*}$.