- #1

mbrmbrg

- 496

- 2

[tex]\int\frac{dx}{a^2-x^2}[/tex]

My Work:

[tex]\frac{1}{a^2-x^2}[/tex]

[tex]=\frac{1}{(a+x)(a-x)}=\frac{A}{a+x}+\frac{B}{a-x}}[/tex]

[tex]1=A(a-x)+B(a+x)[/tex]

If x=a, then [tex]1=2Ba[/tex] so [tex]B=\frac{1}{2a}[/tex]

Thus [tex]1=A(a-x)+\frac{1}{2a}(a+x)[/tex]

if x=0, then [tex]1=Aa+\frac{1}{2}[/tex] so [tex] A=\frac{1}{2a}[/tex]

SO

[tex]\int\frac{dx}{a^2-x^2}[/tex]

[tex]=\int\frac{\frac{1}{2a}}{a+x}+\frac{\frac{1}{2a}}{a-x}dx[/tex]

[tex]=\frac{1}{2a}\int\frac{dx}{a+x}+\frac{1}{2a}\int\frac{dx}{a-x}[/tex]

[tex]=\frac{1}{2a}\ln\mid a+x\mid +\frac{1}{2a}\ln\mid a-x\mid(-1)[/tex]

[tex]=\frac{1}{2a}(\ln\mid a+x\mid -\ln\mid a-x\mid)[/tex]

[tex]=\frac{1}{2a}\ln\mid\frac{a+x}{a-x}\mid +C[/tex]

Table of integrals gives correct answer as

[tex]=\frac{1}{2a}\ln\mid\frac{x+a}{x-a}\mid +C[/tex]

My gut feeling is that I messed up integrating [tex]\int\frac{dx}{a-x}[/tex] but I can't find my error.

Any help would be appreciated.