How do you integrate $(\frac{1}{25 + x^{2}}) d x$ $(1/(25 + x^2) ) dx$ ?

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Answer 1 · 2015-05-18T00:11:05

We can rewrite this as

$\int (\frac{1}{5^{2} + x^{2}}) d x$ $int(1/(5^2+x^2))dx$

By definition, we know that the integration of the inverse of a sum of squares results in a trigonometric function as follows:

$\int (\frac{1}{u^{2} + a^{2}}) d u = (\frac{1}{a}) arctan (\frac{u}{a}) + c$ $int(1/(u^2+a^2))du=(1/a)arctan(u/a)+c$

Applying this formula to your function, we have this:

$\int (\frac{1}{5^{2} + x^{2}}) d x = \frac{arctan (\frac{x}{5})}{5} + c$ $int(1/(5^2+x^2))dx=color(green)((arctan(x/5))/5+c)$

How do you integrate (1/(25 + x^2) ) dx(125+x2)dx(1/(25 + x^2) ) dx?