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Integration

Question 9 of 11

Beginning Integration — The Reverse of Differentiation

Question

Integrate \(\int\left(w + \dfrac{1}{w}\right)\left(w - \dfrac{1}{w}\right)dw\)

Show solution Hide solution Fully worked — 4 steps
  1. Expand the product
    \[ \int\left(w + \frac{1}{w}\right)\left(w - \frac{1}{w}\right)dw \]

    Express the product as a difference of two squares

    \[ = \int\left(w^2 - \frac{1}{w^2}\right)dw \]
  2. Split the integral into a sum

    Use \(\int f(x)\,dx + g(x)\,dx = \int f(x)\,dx + \int g(x)\,dx\)

    \[ = \int w^2\,dw - \int \frac{1}{w^2}\,dw \]
  3. Express in negative exponential form
    \[ = \int w^2\,dw - \int w^{-2}\,dw \]
  4. Apply the rule and simplify

    Use \(\int x^n\,dx = \frac{1}{n+1}x^{n+1} + c\). Simplify

    \[ \int\left(w + \frac{1}{w}\right)\left(w - \frac{1}{w}\right)dw = \frac{w^3}{3} + \frac{1}{w} + c \]

    That is, the integral is \(w\) cubed over 3, plus 1 over \(w\), plus the constant of integration \(c\).