Evaluate $\int{ \ln(\sqrt{x})dx}$ using integration by parts

Question

In Khan Academy practice questions on Integration by Parts, the above question was asked. And the following options were provided:

A)    $x\ln(x) - x + C$

B)    $\frac{1}{2}x\ln(x) - \frac{1}{2}x + C$

C)    $x\ln(\sqrt{x}) - \frac{2}{3}x^{3/2} + C$

D)    $\frac{1}{2}x\ln(\sqrt{x}) - \frac{1}{3}x^{3/2} + C$

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My attempt:

We can rewrite the integral as, $$\int{\ln(\sqrt{x})(1)dx}$$

Using Integration by parts, it can be given as, $$ = \ln(\sqrt{x})\int{1 \, dx} - \int{\frac{d(\ln\sqrt{x})}{dx}\left(\int{1.dx}\right)dx}$$

$$ = \ln(\sqrt{x})\int{1 \, dx} - \int{\frac{1}{\sqrt{x}}\left(\frac{1}{2\sqrt{x}}\right)xdx}$$

$$ = x \ln(\sqrt{x}) - \frac{x}{2} + C$$

But, my answer doesn't matches any of the above options. What I am doing wrong?

Answer 1

You have not done anything wrong. Simply note that $\log(\sqrt x)=\frac12 \log(x)$.

Therefore, your analysis is correct; the answer is $(B)\,\, \frac12x\log(x)-\frac12 x+C$.

Answer 2

Another way: set $\sqrt{x} = t$ to get $\int \log t d (t^2) = 2 \int t \log t dt $ and then proceed integrating by parts