Integral of 1/(x^2 – a^2)

The Mathematician — Mon, 05 Jun 2023 13:00:39 +0000

What is the integral of 1/(x^2 – a^2)?

The integral of

\frac{1}{x^2 - a^2}

\frac{1}{2a}\ln\lvert \frac{x - a}{x + a} \rvert + C

Solution of the integral 1/(x^2 – a^2)

Solution: We want to determine the next integral:

\begin{align*} \int \frac{dx}{x^2 - a^2}. \end{align*}

We will apply an integral technique such that the denominators will be linear functions:

\begin{align*} \frac{1}{x^2 - a^2} &= \frac{1}{(x-a)(x+a)} \\ &= \frac{A}{x - a} + \frac{B}{x + a} \\ &= \frac{Ax + Aa + Bx - Ba}{x^2 - a^2}. \end{align*}

We do have the following set of equalities:

\begin{align*} A + B &= 0 \\ Aa - Ba &= 1. \end{align*}

We see that

A = -B,

and therefore

B = -\frac{1}{2a}

and

A = \frac{1}{2a}

. Substituting everything above, we get:

\begin{align*} \int \frac{1}{x^2 - a^2} dx &= \int \frac{A}{x - a} dx + \int \frac{B}{x + a} dx \\ &= \frac{1}{2a} \int \frac{dx}{x - a} - \frac{1}{2a} \int \frac{dx}{x + a} \\ &= \frac{1}{2a} \ln \lvert x - a \rvert - \frac{1}{2a} \ln \lvert x + a \rvert + C \\ &= \frac{1}{2a} \ln \bigg\lvert \frac{x - a}{x + a} \bigg\rvert + C. \end{align*}

Conclusion

In conclusion, the integral of

\frac{1}{x^2 - a^2}

\frac{1}{2a}\ln \lvert \frac{x - a}{x + a} \rvert + C

, or in mathematical notation:

\begin{equation*} \int \frac{dx}{x^2 - a^2} = \frac{1}{2a} \ln \bigg\lvert \frac{x - a}{x + a} \bigg\rvert + C. \end{equation*}The post Integral of 1/(x^2 – a^2) appeared first on Epsilonify.

Integral of 1/(x^2 + a^2)

The Mathematician — Tue, 30 May 2023 13:00:46 +0000

What is the integral of 1/(x^2 + a^2)?

The integral of

\frac{1}{x^2 + a^2}

\frac{1}{a}\tan^{-1}(\frac{x}{a}) + C

Solution of the integral of 1/(x^2 + a^2)

Solution: While we could straightly use some integral techniques, the reader should remind itself that there is a function that if you take the derivative of that, then we get exactly the solution of

\frac{1}{x^2 + a^2}

. Indeed, we get the following from this article:

\begin{align*} \frac{d}{dx} \frac{1}{a}\tan^{-1}\bigg(\frac{x}{a}\bigg) = \frac{1}{x^2 + a^2}. \end{align*}

Now take the antiderivative of

\frac{1}{x^2 - a^2}

, then we get that the integral of

\frac{1}{x^2 + a^2}

\frac{1}{a}\tan^{-1}(\frac{x}{a}) + C

Conclusion

Therefore, the integral of

\frac{1}{x^2 + a^2}

\frac{1}{a}\tan^{-1}(\frac{x}{a}) + C

, or in mathematical notation:

\begin{align*} \int \frac{dx}{x^2 + a^2} = \frac{1}{a}\tan^{-1}\bigg(\frac{x}{a}\bigg) + C. \end{align*}

The post Integral of 1/(x^2 + a^2) appeared first on Epsilonify.

Integral of 1/(x^2 - a^2) Archives - Epsilonify

Integral of 1/(x^2 – a^2)

What is the integral of 1/(x^2 – a^2)?

Solution of the integral 1/(x^2 – a^2)

Conclusion

Integral of 1/(x^2 + a^2)

What is the integral of 1/(x^2 + a^2)?

Solution of the integral of 1/(x^2 + a^2)

Conclusion