Question

Evaluate the definite integral
1
∫1/[(x+1)(x^2-4)] dx
0

Answer 1

Partial Fractions

Answer 2

First :
We can write :
\[\frac{1}{(x+1)(x^2-4)}=\frac{a}{x+1}+\frac{b}{x-2}+\frac{c}{x+2}
\\=\frac{a(x-2)+b(x+1)}{(x+1)(x-2)}+\frac{c}{x+2}
\\=\frac{(a+b)x+b-2a}{(x+1)(x-2)}+\frac{c}{x+2}
\\=\frac{(a+b)x(x+2)+b(x+2)-2a(x+2)+c(x+1)(x+2)}{(x+1)(x-2)(x+2)}
\\=\frac{(a+b+c)x^2+(3b-c)x-2(2a+-b-c)}{(x+1)(x^2-4)}
\]
So :
\[a+b+c=0\\3b-c=0\\-2(2a-b-c)=1\]
Then :
\[a=-1/6~~~b=1/24~~~c=1/8\]
Then :
\[\int_0^1\frac{1}{(x+1)(x^2-4)}dx=[a\ln(x+1)+b\ln(2-x)+c\ln(x+2)]_0^1\\
=a\ln2-b\ln2+c\ln3-c\ln2=(a-b-c)\ln2+c\ln 3\]

Answer 3

$$\frac1{(x+1)(x^2-4)}=\frac1{(x-2)(x+1)(x+2)}=\frac{A}{x-2}+\frac{B}{x+1}+\frac{C}{x+2}\\A(x+1)(x+2)+B(x-2)(x+2)+C(x-2)(x+1)=1$$We can plug in \(x=-1\) and watch what happens:$$A(-1+1)(-1+2)+B(-1-2)(-1+2)+C(-1-2)(-1+1)=1\\0+(-3)(1)B+0=1\\-3B=1\\B=-\frac13$$Similarly, plugging in \(x=-2\) we eliminate the terms with \((x+2)\) factors:$$C(-2-2)(-2+1)=1\\(-4)(-1)C=1\\4C=1\\C=\frac14$$Doing something similar for \(A\) we substitute \(x=2\):$$(2+1)(2+2)A=1\\(3)(4)A=1\\12A=1\\A=\frac1{12}$$... so now we've completed our partial fraction decomposition:$$\frac1{(x+1)(x^2-4)}=\frac1{12}\times\frac1{x-2}-\frac13\times\frac1{x+1}+\frac14\times\frac1{x+2}$$Now we can integrate term by term:$$\int\frac1{(x+1)(x^2-4)}dx=\frac1{12}\log(x-2)-\frac13\log(x+1)+\frac14\log(x+2)+C$$Now we evaluate at \(x=1\) and \(x=0\) and find their difference:$$\frac12\color{red}{\log(-1)}+\dots$$oops! We need to change our partial fraction decomposition a little bit:$$\frac1{(x+1)(x^2-4)}=-\frac1{12}\times\frac1{2-x}-\frac13\times\frac1{x+1}+\frac14\times\frac1{x+2}$$We merely rewrote \(1/12\times1/(x-2)=-1/12\times1/(2-x)\). Now we reintegrate:$$\int\frac1{(x+1)(x^2-4)}dx=\frac1{12}\log(2-x)-\frac13\log(x+1)+\frac14\log(x+2)+C$$Try evaluating now:$$\frac1{12}\log1-\frac13\log2+\frac14\log3-\frac1{12}\log2+\frac13\log1-\frac14\log2$$All of the \(\log1\) terms vanish since \(\log1=0\):$$-\frac13\log2-\frac1{12}\log2-\frac14\log2+\frac14\log3=-\frac23\log2+\frac14\log3$$