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Evaluate $$\int \frac{x^2}{16x^2+1}\,\mathrm dx.$$

Apparently we are supposed to let $x = \dfrac14u$ and that somehow gives $$\frac1{64}\int \frac{u^2}{u^2+1}\,\mathrm du.$$

I don't get how this happens and how do you come up with $x = \dfrac14u$??

After this, I understand the rest of trying to get the integral to look like the integral of $\arctan$. It's just the above part that I don't get. Can anybody help me?

Tianlalu
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Yanni Papaioannou
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1 Answers1

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The idea here is that $16x^2$ is basically $(4x)^2$ so the actual substitution is $4x = u$.

If you had $25$ instead of $16$, then your substitution would be $x = 1/5*u$ or $5x = u$ since you'd have $(5x)^2$ instead.

It is all about patterns really,

Makina
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