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#