Question #07873

1 Answer
Mar 4, 2016

Answer:

#int(1)/(x+sqrtx)dx = 2 ln(sqrt(x) + 1) + C#, with #x>= 0# and where #C# is the constant of integration.

Explanation:

We want to compute the antiderivative of #int(1)/(x+sqrtx)dx#.
Since we have #sqrt(x)#, we'll assume #x>=0#.

We will use this change of variable :
#x = phi(t) = t^2#

#int(1)/(x+sqrtx)dx# = #int(1)/(t^2+sqrt(t^2))*dot ((t^2))dt#

#= int(2t)/(t^2+t)dt = 2 int 1/(t+1) dt = 2 ln(|t+1|) + C.#

Since #x = t^2#,# t = sqrt(x)# for #x >= 0# (which was supposed above).

So #int(1)/(x+sqrtx)dx = 2 ln(|sqrt(x) + 1|) + C = 2 ln(sqrt(x) + 1) + C#, where #C# is the constant of integration.