What is #f(x) = int x+5sqrt(x^2+1) dx# if #f(2) = 7 #?

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Answer 1 · 2017-02-20T20:58:20

#f(x)=1/2(x^2+5xsqrt(x^2+1)+5ln|x+sqrt(x^2+1)|)-98/10#

#f(x)=intx+5sqrt(x^2+1)# #dx#

Let #x = tanu#

#dx/(du)=sec^2u#

#dx=sec^2u# #du#

#cosu=1/sqrt(1+x^2)#

#sqrt(1+x^2)=1/cosu=secu#

#intf(x)# #dx=intx+5intsecusec^2u# #du#

#intx# #dx=1/2x^2+C#

#5intsecusec^2u# #du=5(secutanu-intsecutan^2u# #du)#

#=5 secu tanu-5 int sec^3u# #du+5intsec# #du#

#10intsecusec^2u# #du=5secutanu+5ln|secu+tanu|+C#

#5intsecusec^2u# #du=1/2(5secutanu+5ln|secu+tanu|)+C#

#intf(x)# #dx=1/2(x^2+5xsqrt(x^2+1)+5ln|x+sqrt(x^2+1)|)+C#

#f(2)=7=1/2(4+10sqrt(5)+5ln|2+sqrt5|)+C#

#C=-98/10# to #2# dp.

#f(x)=1/2(x^2+5xsqrt(x^2+1)+5ln|x+sqrt(x^2+1)|)-98/10#

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