Integration – Partial Fractions – One Problem at a Time

Problem 234: Evaluate the integral

(1) $\begin{equation*} \int \frac{x^2}{x^2 + 64} \ dx \end{equation*}$

Solution: First, we are going to use long division to rewrite our integral. That is, using long division,

(2) $\begin{equation*} \frac{x^2}{x^2 + 64} = 1 -\frac{64}{x^2 + 64}.\end{equation*}$

One can now use this to rewrite our integral and integral. Thus,

(3) $\begin{align*} \frac{x^2}{x^2 + 64} &= 1 -\frac{64}{x^2 + 64} \\ \int \frac{x^2}{x^2 + 64} \ dx & = \int 1 - 64\frac{1}{x^2 + 64} \ dx \\ & = \int 1 \ dx - \int \frac{64}{x^2 + 64} \ dx \\ & = x - 64 \int \frac{1}{(x)^2 + (8)^2} \ dx \\ & = x - 64 \cdot \frac{1}{8} \cdot \tan^{-1}\left(\frac{x}{8} \right) + C \\ & = x - 8\tan^{-1}\left( \frac{x}{8}\right)+C\end{align*}$

That is,

(4) $\begin{equation*}\int \frac{x^2}{x^2 + 64} \ dx = x - 8\tan^{-1}\left( \frac{x}{8}\right)+C. \end{equation*}$

Integration – Partial Fractions

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