What is the integral of #x*sqrt(25+x^2)#?

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Answer 1 · 2015-07-29T17:34:59

Notice how #sqrt(25 + x^2) prop sqrt(a^2 + x^2)#, which implies #x = atantheta# with #a = 5#.

If we let:

#x = 5tantheta#
#dx = 5sec^2thetad theta#
#sqrt(25 + x^2) = sqrt(5^2 + 5^2tan^2theta) = 5sectheta#

Then we get:

#= int 5tantheta*5sectheta*5sec^2thetad theta#

#= 125int tanthetasecthetasec^2thetad theta#

Then, if we let:

#u = sectheta#
#du = secthetatanthetad theta#

we get:

#= 125int u^2du#

#= 125(u^3/3)#

#= 125((sec^3theta)/3)#

#= ((5^3sec^3theta)/3)#

#= ((5sectheta)^3)/3#

#= ((sqrt(25 + x^2))^3)/3#

#= color(blue)(((25 + x^2)^("3/2"))/3 + C)#