Question

consider the indefinite integral $int \frac{x}{(x^{2}-7)^{2}}dx$: this can be transformed into a basic integral by letting $u=square$ and $du=square dx$ performing the substitution yields the integral $int square du$ question help: $square$ video

Explanation:

Step1: Choose substitution variable

Let \(u = x^{2}-7\).

Step2: Differentiate \(u\)

Differentiate \(u\) with respect to \(x\). Using the power rule \(\frac{d}{dx}(x^{n})=nx^{n - 1}\), we have \(\frac{du}{dx}=2x\), so \(du = 2x dx\), and \(x dx=\frac{1}{2}du\).

Step3: Substitute into the integral

The original integral \(\int\frac{x}{(x^{2}-7)^{2}}dx\). Substitute \(u = x^{2}-7\) and \(x dx=\frac{1}{2}du\). We get \(\int\frac{1}{u^{2}}\cdot\frac{1}{2}du=\frac{1}{2}\int u^{- 2}du\).

Answer:

\(u = x^{2}-7\), \(du = 2x dx\), \(\frac{1}{2}\int u^{-2}du\)