Can I use standard integrals not found on the reference sheet? (1 Viewer)

Poly · Feb 11, 2026

For example:

WeiWeiMan · Feb 11, 2026

Poly said:
For example: View attachment 51616

no

u gotta split this one up into partial fractions

like:
$\begin{align*} \int \frac{dx}{x^2-a^2} &= \frac{1}{2a} \int \frac{2a}{x^2-a^2} dx \\ &= \frac{1}{2a} \int \frac{(x+a)-(x-a)}{(x-a)(x+a)} dx \\ &= \frac{1}{2a} \int \frac{1}{x-a} - \frac{1}{x+a} dx \\ &= \frac{1}{2a}\left(\ln|x-a| - \ln|x+a| \right) + C \\ &= \frac{1}{2a} \ln\left|\frac{x-a}{x+a}\right| + C \end{align*}$

coolcat6778 · Feb 15, 2026

WeiWeiMan said:
no

u gotta split this one up into partial fractions

like:
$\begin{align*} \int \frac{dx}{x^2-a^2} &= \frac{1}{2a} \int \frac{2a}{x^2-a^2} dx \\ &= \frac{1}{2a} \int \frac{(x+a)-(x-a)}{(x-a)(x+a)} dx \\ &= \frac{1}{2a} \int \frac{1}{x-a} - \frac{1}{x+a} dx \\ &= \frac{1}{2a}\left(\ln|x-a| - \ln|x+a| \right) + C \\ &= \frac{1}{2a} \ln\left|\frac{x-a}{x+a}\right| + C \end{align*}$

where tf did u find the time to learn latex

WeiWeiMan · Feb 15, 2026

coolcat6778 said:
where tf did u find the time to learn latex

idk
latex very easy to learn tho

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Can I use standard integrals not found on the reference sheet? (1 Viewer)

Poly

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WeiWeiMan

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coolcat6778

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