How do I evaluate the integral (Calculus 2)

Integral 1 / ((sqrt(x))(1+x))dx

It tells me to change x=(sqrt(x)^2) to make things easier. But, for me it just made things a little bit harder. I can do it without changing it. Can any help? Thanks

Hello,

according to the hint, rewrite the integral as:

∫ {1 /[1 + (√x)²]} (1/√x) dx =

let:

√x = u

differentiate both sides:

d[x^(1/2)] = du

(1/2) x^[(1/2) -1] dx = du

(1/2)x^(-1/2) dx = du

(1/2) (1/√x) dx = du

(1/√x) dx = 2 du

then, substituting:

∫ {1 /[1 + (√x)²]} (1/√x) dx = ∫ [1 /(1 + u²)] 2 du =

(pulling the constant out)

2 ∫ [1 /(1 + u²)] du =

2arctan u + C

being u = √x, the answer is:


∫ {1 /[(√x)(1 + x)]} dx = 2arctan(√x) + C


I hope it's helpful