What is #int_0^4 x/(x^2+9)^(1/2) dx# ?

1 Answer
George C.
Feb 15, 2017

#int_0^4 x/(x^2+9)^(1/2) dx = 2#

Explanation:

Note that:

#d/(dx) (x^2 + 9) = 2x#

So:

#int_0^4 x/(x^2+9)^(1/2) dx = int_0^4 1/2(d/(dx)(x^2+9))(x^2+9)^(-1/2) dx#

#color(white)(int_0^4 x/(x^2+9)^(1/2) dx) = [(x^2+9)^(1/2)]_0^4#
#color(white)(int_0^4 x/(x^2+9)^(1/2) dx) = (color(blue)(4)^2+9)^(1/2)-(color(blue)(0)^2+9)^(1/2)#
#color(white)(int_0^4 x/(x^2+9)^(1/2) dx) = (16+9)^(1/2)-(9)^(1/2)#
#color(white)(int_0^4 x/(x^2+9)^(1/2) dx) = 5-3#
#color(white)(int_0^4 x/(x^2+9)^(1/2) dx) = 2#

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