What is $int 1 / (25 + x^2 ) dx$ ?

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Answer 1 · 2016-03-05T16:09:16

$1/5Tan^-1(x/5)+c$

$int1/(25+x^2)dx$

$=intdx/(5^2+x^2)$

$=I$

We use the rule:

$intdx/(a^2+x^2)=1/aTan^-1(x/a)+c$

Here $a=5$

$implies I=1/5Tan^-1(x/5)+c$

Answer 2 · 2016-09-16T03:20:57

$1/5arctan(x/5)+C$

Alternatively, apply the substitution $x=5tan(y)$ . Note that this implies that $dx=5sec^2(y)dy$ .

$intdx/(25+x^2)=int(5sec^2(y)dy)/(25+25tan^2(y))=1/5int(sec^2(y)dy)/(1+tan^2(y))$

Using the identity $1+tan^2(y)=sec^2(y)$ :

$=1/5int(sec^2(y)dy)/(sec^2(y))=1/5intdy=1/5y+C$

From $x=5tan(y)$ we see that $y=arctan(x/5)$ :

$=1/5arctan(x/5)+C$

What is int 1 / (25 + x^2 ) dxint 1 / (25 + x^2 ) dx?