How do you integrate #int x^3lnx# by integration by parts method?

Question

William Abdalla · Accepted Answer

#intx^3ln(x)dx=(x^4(4ln(x)-1))/16+C#

Integration by parts takes the form:

#intudv=uv-intvdu#

So, for the integral #intx^3ln(x)dx#, we see this is #intudv# and let:

#{(u=ln(x)" "=>" "du=1/xdx),(dv=x^3dx" "=>" "v=x^4/4):}#

Thus:

#intx^3ln(x)dx=uv-intvdu=(x^4ln(x))/4-intx^4/4 1/xdx#

Simplifying the integral:

#intx^3ln(x)dx=(x^4ln(x))/4-1/4intx^3dx#

#intx^3ln(x)dx=(x^4ln(x))/4-1/4x^4/4+C#

#intx^3ln(x)dx=(x^4ln(x))/4-x^4/16+C#

#intx^3ln(x)dx=(x^4(4ln(x)-1))/16+C#

Christopher Agresta · Answer

# int x^3lnxdx=1/4x^4lnx - x^4/16 + C #

The formula for integration by parts is: # intu(dv)/dxdx = uv - intv(du)/dxdx #

Essentially we would like to find one function that simplifies when differentiated, and one that simplifies when integrated (or is at least integrable).

Let # {(u=lnx, => ,(du)/dx=1/x),((dv)/dx=x^3,=>,v =x^4/4 ):}#

So IBP gives:

# int (lnx)(x^3)dx=(lnx)(x^4/4) - int (x^4/4)(1/x)dx # # :. int x^3lnxdx=1/4x^4lnx - 1/4int x^3dx # # :. int x^3lnxdx=1/4x^4lnx - 1/4(x^4/4) + C # # :. int x^3lnxdx=1/4x^4lnx - x^4/16 + C #

Adam Adkins · Answer

undefined To integrate $ \int x^3 \ln(x) $ using integration by parts, you would apply the formula:

$$ \int u \, dv = uv - \int v \, du $$

Let $ u = \ln(x) $ and $ dv = x^3 \, dx $. Then, differentiate $ u $ to get $ du $ and integrate $ dv $ to get $ v $.

$$ du = \frac{1}{x} \, dx $$
$$ v = \frac{1}{4}x^4 $$

Now, apply the formula:

$$ \int x^3 \ln(x) \, dx = \frac{1}{4}x^4 \ln(x) - \int \frac{1}{4}x^4 \cdot \frac{1}{x} \, dx $$

Simplify and solve the integral:

$$ \int x^3 \ln(x) \, dx = \frac{1}{4}x^4 \ln(x) - \frac{1}{4} \int x^3 \, dx $$

$$ = \frac{1}{4}x^4 \ln(x) - \frac{1}{16}x^4 + C $$

Therefore, $ \int x^3 \ln(x) \, dx = \frac{1}{4}x^4 \ln(x) - \frac{1}{16}x^4 + C $, where $ C $ is the constant of integration.