Integral of x/(x^2 – a^2)

The Mathematician — Fri, 26 May 2023 13:00:52 +0000

What is the integral of x/(x^2 – a^2)

The integral of

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

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

Solution: Before we determine the integral of

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

, lets us recall what we need to show exactly:

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

We use the substitution method where

u = x^2 - a^2

such that we get the derivative

du = 2xdx \iff xdx = \frac{1}{2}du

. Now we will implement that in the integral above, and therefore, we get the desired solution:

\begin{align*} \int \frac{x}{x^2 - a^2} dx &= \int \frac{\frac{1}{2}du}{u} \\ &= \frac{1}{2} \int \frac{du}{u} \\ &= \frac{1}{2} \ln \lvert u \rvert + C \\ &= \frac{1}{2} \ln \lvert x^2 - a^2 \rvert + C \end{align*}

The detailed solution above shows us that the integral of

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

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

The Mathematician — Mon, 22 May 2023 13:00:26 +0000

The integral of

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

\frac{1}{2}\ln(x^2 + a^2) + C

We want to determine the integral of

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

, i.e.,

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

We will use the substitution method. Let

u = x^2 + a^2

. Then

du = 2xdx \iff \frac{1}{2}du = xdx

. So we get the following integral:

\begin{align*} \int \frac{x}{x^2 + a^2} dx &= \int \frac{\frac{1}{2}du}{u} \\ &= \frac{1}{2} \int \frac{du}{u} \\ &= \frac{1}{2}\ln \lvert u \rvert + C \\ &= \frac{1}{2}\ln (x^2 + a^2) + C. \end{align*}

So, the integral of

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

\frac{1}{2}\ln(x^2 + a^2) + C