Solving Integration by Parts Problem: \int\frac{dx}{a^2-x^2}

• mbrmbrg
In summary, the conversation discussed the problem of integrating \frac{dx}{a^2-x^2}. The individual provided their work, which involved breaking the fraction into smaller parts and finding the values for A and B. They then used these values to solve the integral and ended up with the equation \frac{1}{2a}\ln\mid\frac{x+a}{x-a}\mid +C. The individual expressed uncertainty about their solution but it was confirmed to be correct in the end.
mbrmbrg
Problem:
$$\int\frac{dx}{a^2-x^2}$$
My Work:
$$\frac{1}{a^2-x^2}$$
$$=\frac{1}{(a+x)(a-x)}=\frac{A}{a+x}+\frac{B}{a-x}}$$
$$1=A(a-x)+B(a+x)$$
If x=a, then $$1=2Ba$$ so $$B=\frac{1}{2a}$$
Thus $$1=A(a-x)+\frac{1}{2a}(a+x)$$
if x=0, then $$1=Aa+\frac{1}{2}$$ so $$A=\frac{1}{2a}$$
SO
$$\int\frac{dx}{a^2-x^2}$$
$$=\int\frac{\frac{1}{2a}}{a+x}+\frac{\frac{1}{2a}}{a-x}dx$$
$$=\frac{1}{2a}\int\frac{dx}{a+x}+\frac{1}{2a}\int\frac{dx}{a-x}$$
$$=\frac{1}{2a}\ln\mid a+x\mid +\frac{1}{2a}\ln\mid a-x\mid(-1)$$
$$=\frac{1}{2a}(\ln\mid a+x\mid -\ln\mid a-x\mid)$$
$$=\frac{1}{2a}\ln\mid\frac{a+x}{a-x}\mid +C$$

Table of integrals gives correct answer as
$$=\frac{1}{2a}\ln\mid\frac{x+a}{x-a}\mid +C$$

My gut feeling is that I messed up integrating $$\int\frac{dx}{a-x}$$ but I can't find my error.
Any help would be appreciated.

Its the same thing, because its in absolute values.

What is integration by parts?

Integration by parts is a technique used to solve integrals by breaking them down into smaller, simpler integrals. It involves using the product rule from calculus to rewrite the original integral into a more manageable form.

How do I know when to use integration by parts?

Integration by parts is typically used when the integral involves a product of two functions, one of which can be differentiated and the other can be integrated. In the case of \int\frac{dx}{a^2-x^2}, we can use integration by parts because the integral involves the product of \frac{1}{a^2-x^2} and 1, which can be differentiated.

What are the steps for solving integration by parts?

The general steps for solving an integral using integration by parts are:

1. Identify the functions u and dv in the original integral.
2. Use the product rule to rewrite the integral as \int u dv = uv - \int v du.
3. Solve the simpler integral \int v du using traditional integration techniques.
4. Substitute the values for u and v back into the original integral to get the final solution.

Can integration by parts be used for any integral?

No, integration by parts can only be used for certain types of integrals, specifically those involving a product of two functions, one of which can be differentiated and the other can be integrated. Other techniques, such as substitution or partial fractions, may be more appropriate for other types of integrals.

How can I check if my answer is correct?

You can check your answer by differentiating it and seeing if it matches the original integrand. In the case of \int\frac{dx}{a^2-x^2}, you can differentiate your final solution and see if it matches \frac{1}{a^2-x^2}. You can also use online integrator tools to verify your answer.

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