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How to integrate 1/(64x^2+100)?

1 Answer

Jrab · Ananda Dasgupta

May 22, 2018

$\frac{1}{80} arctan (\frac{4 x}{5}) + c$ $1/80arctan((4x)/5)+c$

Explanation:

The integral is $\int \frac{d x}{64 x^{2} + 100}$ $int(dx)/(64x^2+100)$

divide the denominator by 64; $\frac{1}{64} \int \frac{d x}{x^{2} + (\frac{25}{16})}$ $1/64intdx/(x^2+(25/16)$

simplify; $\frac{1}{64} \int \frac{d x}{x^{2} + {(\frac{5}{4})}^{2}}$ $1/64intdx/(x^2+(5/4)^2$

use the formula $\int \frac{d x}{a^{2} + x^{2}} = \frac{1}{a} arctan (\frac{x}{a}) + c$ $intdx/(a^2+x^2)=1/aarctan(x/a)+c$

plug in $(\frac{1}{80}) arctan (\frac{4 x}{5}) + c$ $(1/80)arctan((4x)/5) + c$

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