Question

7–58. Improper integrals Evaluate the following integrals or state that they diverge.56. ∫ (from 0 to 1) 1/(x + √x) dx

Accepted Answer

First, analyze the integrand $$ \frac{1}{x + \sqrt{x}}  to $$check for any points of discontinuity or where the function might be undefined on the interval $$[0,1]. $$Notice that at $$x=0$$, both $$x$$ and $$\sqrt{x}$$ are zero, so the denominator approaches zero, indicating a potential improper integral at the lower limit. Rewrite the integrand to a simpler form to make it easier to integrate. Factor the denominator as $$ x + \sqrt{x} = \sqrt{x}(\sqrt{x} + 1) $$, so the integrand becomes $$ \frac{1}{\sqrt{x}(\sqrt{x} + 1)} . $$Use a substitution to simplify the integral. Let $$ t = \sqrt{x} $$, which implies $$ x = t^2 $$ and $$ dx = 2t \, dt . $$Change the limits accordingly: when $$ x=0 $$, $$ t=0 $$; when $$ x=1 $$, $$ t=1 . $$Rewrite the integral in terms of $$ t $$: $$ \int_0^1 \frac{1}{t(t+1)} \cdot 2t \, dt = \int_0^1 \frac{2t}{t(t+1)} \, dt = \int_0^1 \frac{2}{t+1} \, dt. $$ This simplifies the integral significantly. Now, integrate $$ \int_0^1 \frac{2}{t+1} \, dt  by $$recognizing it as a standard logarithmic integral. After integrating, evaluate the resulting expression at the limits $$ t=0 $$ and $$ t=1  to $$find the value of the improper integral.

7–58. Improper integrals Evaluate the following integrals or state that they diverge.
56. ∫ (from 0 to 1) 1/(x + √x) dx

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