How do you integrate #intx^(2/3) *ln x# from 1 to 4?

Answer 1
#int x^r lnx dx# is a 'standard' question.
For #r!= -1# integrate by parts.
We don't (most of us) know the integral of #lnx#, but we do know its derivative, so Let #u = lnx# and #dv = x^r dx# (in this question #dv = x^(2/3) dx#

With these choices, we get:

#du = 1/x dx# and #v = int x^r dx = x^(r+1)/(r+1)# (Here: #3/5x^(5/3)#)
#int u dv = uv - int vdu = 3/5 x^(5/3) lnx - 3/5 int x^(5/3) * 1/x dx#
#= 3/5 x^(5/3) lnx - 3/5 int x^(2/3) dx# (You'll always get the same integral fo this kind of problem.)
#= 3/5 x^(5/3) lnx - 3/5 *3/5 x^(5/3)+C# Or for your definite integral:
#= [3/5 x^(5/3) lnx - 9/25 x^(5/3)]_1^4#
Now do the arithmetic. (remember that #ln1 = 0#)
Sign up to view the whole answer

By signing up, you agree to our Terms of Service and Privacy Policy

Sign up with email
Answer 2

To integrate the function ( \int x^{\frac{2}{3}} \ln(x) ) from 1 to 4, use integration by parts.

Let ( u = \ln(x) ) and ( dv = x^{\frac{2}{3}} dx ). Then, ( du = \frac{1}{x} dx ) and ( v = \frac{3}{5}x^{\frac{5}{3}} ).

Now apply the integration by parts formula:

[ \int u , dv = uv - \int v , du ]

Substitute the values:

[ \int x^{\frac{2}{3}} \ln(x) , dx = \frac{3}{5}x^{\frac{5}{3}} \ln(x) - \int \frac{3}{5}x^{\frac{5}{3}} \cdot \frac{1}{x} , dx ]

Simplify the integral:

[ = \frac{3}{5}x^{\frac{5}{3}} \ln(x) - \frac{3}{5} \int x^{\frac{2}{3}} , dx ]

Now integrate the remaining integral:

[ = \frac{3}{5}x^{\frac{5}{3}} \ln(x) - \frac{3}{5} \left( \frac{3}{5}x^{\frac{5}{3}} \right) + C ]

Evaluate the integral from 1 to 4:

[ = \left[ \frac{3}{5}x^{\frac{5}{3}} \ln(x) - \frac{9}{25}x^{\frac{5}{3}} \right]_1^4 ]

[ = \left( \frac{3}{5} \cdot 4^{\frac{5}{3}} \ln(4) - \frac{9}{25} \cdot 4^{\frac{5}{3}} \right) - \left( \frac{3}{5} \cdot 1^{\frac{5}{3}} \ln(1) - \frac{9}{25} \cdot 1^{\frac{5}{3}} \right) ]

[ = \left( \frac{3}{5} \cdot 4^{\frac{5}{3}} \ln(4) - \frac{9}{25} \cdot 4^{\frac{5}{3}} \right) - \left( 0 - \frac{9}{25} \right) ]

[ = \frac{3}{5} \cdot 4^{\frac{5}{3}} \ln(4) - \frac{9}{25} \cdot 4^{\frac{5}{3}} + \frac{9}{25} ]

[ ≈ 9.0407 ]

So, the value of the integral of ( \int_{1}^{4} x^{\frac{2}{3}} \ln(x) , dx ) is approximately 9.0407.

Sign up to view the whole answer

By signing up, you agree to our Terms of Service and Privacy Policy

Sign up with email
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.

Not the question you need?

Drag image here or click to upload

Or press Ctrl + V to paste
Answer Background
HIX Tutor
Solve ANY homework problem with a smart AI
  • 98% accuracy study help
  • Covers math, physics, chemistry, biology, and more
  • Step-by-step, in-depth guides
  • Readily available 24/7