Jean Blumer

2021-12-19

Evaluate the integral.
${\int }_{0}^{1}\frac{2\mathrm{ln}\left(x\right)}{\sqrt{x}}dx$

turtletalk75

Expert

Step 1
Consider the provided integral,
${\int }_{0}^{1}\frac{2\mathrm{ln}\left(x\right)}{\sqrt{x}}dx$
Evaluate the above integral.
Step 2
Now, solve the above integral using by part formula.
${\int }_{0}^{1}\frac{2\mathrm{ln}\left(x\right)}{\sqrt{x}}dx=2\cdot {\int }_{0}^{1}\frac{\mathrm{ln}\left(x\right)}{\sqrt{x}}dx$
$=2{\left[2\sqrt{x}\mathrm{ln}\left(x\right)-\int \frac{2}{\sqrt{x}}dx\right]}_{0}^{1}$
$=2{\left[2\sqrt{x}\mathrm{ln}\left(x\right)-4\sqrt{x}\right]}_{0}^{1}$
=2(-4)
=-8

sirpsta3u

Expert

$\int \frac{2\mathrm{ln}\left(x\right)}{\sqrt{x}}dx$
Lets

nick1337

Expert

The integration process can be simplified by making a change of variables:
$t=\sqrt{x},x={t}^{2},dx=2\ast t\ast dt$
Then the original integral can be written as follows: The
$\int 4\ast \mathrm{ln}\left({t}^{2}\right)\ast dt$
$\int 4\ast \mathrm{ln}\left({x}^{2}\right)\ast dx$
formula for integration by parts:
$\int U\left(x\right)\ast dV\left(x\right)=U\left(x\right)\ast V\left(x\right)-\int V\left(x\right)\ast dU\left(x\right)$
Put
$U=4\mathrm{ln}\left({x}^{2}\right)$
dV=dx
Then:
$dU=\frac{8}{x}\ast dx$
V=x
Therefore:
$\int 4\ast \mathrm{ln}\left({x}^{2}\right)\ast dx=4\ast x\ast \mathrm{ln}\left({x}^{2}\right)-\int 8\ast dx$
Find the integral
$\int 8\ast dx=8\ast x$
$\int 4\ast \mathrm{ln}\left({x}^{2}\right)=4\ast x\ast \mathrm{ln}\left({x}^{2}\right)-8\ast x+C$
or
$\int 4\ast \mathrm{ln}\left({x}^{2}\right)=4\ast x\ast \left(\mathrm{ln}\left({x}^{2}\right)-2\right)+C$
To write the final answer, it remains to substitute sqrt (x) instead of t.
$4\sqrt{x}\left(\mathrm{ln}\left(x\right)-2\right)+C$
Let's calculate a definite integral:
${\int }_{0}^{1}\frac{2\ast \mathrm{ln}\left(x\right)}{\sqrt{x}}\ast dx=\left(4\sqrt{x}\ast \left(\mathrm{ln}\left(x\right)-2\right)\right){|}_{0}^{1}$
F(1)=-8
F(0)=
I=-8-(0)=-8

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