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I want to solve $\displaystyle \quad \int\frac{dz}{1+5z^2}$

and Wolfram says I need to use the substitution with inverse tan, taking $u=\sqrt{5}z$, giving me $$\frac{1}{\sqrt{5}}\int\frac{du}{1+u^2}.$$

Surely this is incorrect as the 1 in the demoninator would now be $\sqrt{5}$ too. What don't I understand?

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  • $\begingroup$ After substituting, you get $\int {1\over 1+u^2}\cdot\underbrace{{1\over\sqrt5}\,du}_{dz}$. You then factor $1/\sqrt 5$ out of the integral. The denominator of ${1\over 1+u^2}$ does not come into play here. $\endgroup$ Jan 15 '13 at 16:32
  • $\begingroup$ Thanks, that helps me see it a lot more clearly. Thank you to everyone. $\endgroup$
    – Ian
    Jan 15 '13 at 16:34
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If $\;u = \sqrt{5}z\;$ then $\;du = \sqrt{5}\, dz \;\implies\; \color{red}{\bf{dz = \dfrac{1}{\sqrt5}} \,du}\;$ and

$$\int\frac{1}{1+5z^2}\,dz = \int \frac{1}{1 + (\sqrt{5}z)^2} \,\color{red}{\bf dz}= \int\frac{1}{1+u^2} \cdot \color{red}{\bf{\frac{1}{\sqrt{5}}\,du}}\; = \;\dfrac{1}{\sqrt{5}} \int \dfrac{1}{1+u^2}\,du \;= \;\;?$$

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$du=\sqrt 5 dz$, you have to account for this when substituting

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  • $\begingroup$ So they cancel out and the 1 remains as 1 right? $\endgroup$
    – Ian
    Jan 15 '13 at 16:29
  • $\begingroup$ exactly, but I would recommend doing the integrals on paper until you are comfortable with them $\endgroup$
    – Sean D
    Jan 15 '13 at 16:31

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