#int_0^1 1/(x^2+1)# #=# ?

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Answer 1 · 2017-12-14T21:21:19

# pi/4 #

First thing to sonsider is:

#int 1/(x^2+1) dx #

Make a substitution:

# x = tantheta #

#=> dx = sec^2 theta # #d theta #

#=> int 1 / (tan^2 theta +1 ) sec^ 2 ## d theta #

We must use the identity:

#1 + tan^2 x = sec^2 x #

#=> int sec^2 theta / ( sec^2 theta ) d theta #

#=> int d theta #

#=> theta + c #

Use: #tan^(-1) x = theta #

# = tan^(-1) x + c #

Now considering limits from #0# to #1#

#=> tan^(-1) (1) - tan^(-1)( 0) #

#=> pi/4 - 0 #

#=> pi/4 #

Answer 2 · 2017-12-14T21:24:03

#pi/4#

#int_0^1 1/(x^2+1)dx#

we do this by substitution

#x=tanu#

#=>dx=sec^2udu#

change of limits

#x=0=>u=tan^(-1)0=0#

#x=1=>u=tan^(-1)1=pi/4#

the integral becomes

#int_0^(pi/4)1/cancel(tan^2u+1)xxcancel(sec^2u)du#

#=int_0^(pi/4)du=[u]_0^(pi/4)#

#=pi/4=0=pi/4#