How do you integrate #int x^(1/2)*ln(x) # using integration by parts?

Answer 1

#2/9 x^[3/2)(3ln(x)-2) +C#

Integration by part : #intf(x)g'(x)dx = f(x)g(x)-intf'(x)g(x)dx#
Given #int root()x*ln(x)dx#
Let #f(x)= lnx# #color(blue)(f'(x)= 1/x dx# #" " " " " "(1)#
Let #g'(x)= x^(1/2)dx hArr g(x) = int x^(1/2)dx =>color(red)( g(x)= 2/3 x^(3/2)# # " " " " " " (2)#
#int root()x*ln(x)dx# = #color(red)(2/3 x^(3/2))*ln(x)- intcolor(red)( " " 2/3 x^(3/2)*color(blue)((1/x)dx)#
#2/3 x^(3/2) ln(x) - 2/3 intx^(1/2)dx#
#2/3 x^(3/2) ln(x) - 2/3*2/3(x^(3/2)) + C#
#2/3 x^(3/2) ln(x) - 4/9(x^(3/2)) + C#
Can be simplify to #2/9 x^[3/2)(3ln(x)-2) +C#
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Answer 2

To integrate ( \int x^{1/2} \cdot \ln(x) ) using integration by parts, you'll want to choose ( u = \ln(x) ) and ( dv = x^{1/2} , dx ). Then, differentiate ( u ) to find ( du ) and integrate ( dv ) to find ( v ). Apply the integration by parts formula ( \int u , dv = uv - \int v , du ). Finally, substitute the values back into the formula to find the integral.

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Answer from HIX Tutor

When evaluating a one-sided limit, you need to be careful when a quantity is approaching zero since its sign is different depending on which way it is approaching zero from. Let us look at some examples.

When evaluating a one-sided limit, you need to be careful when a quantity is approaching zero since its sign is different depending on which way it is approaching zero from. Let us look at some examples.

When evaluating a one-sided limit, you need to be careful when a quantity is approaching zero since its sign is different depending on which way it is approaching zero from. Let us look at some examples.

When evaluating a one-sided limit, you need to be careful when a quantity is approaching zero since its sign is different depending on which way it is approaching zero from. Let us look at some examples.

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