How do you integrate #int 3 xln x^3 dx # using integration by parts?

Question

How do you integrate #int  3 xln x^3  dx  # using integration by parts?

Emily Ammerman · Accepted Answer

#int3xln(x^3)dx=(9x^2lnx)/2-9/4x^2+C#

We're going to want to get rid of the #x^3#. Recalling that #ln(x^a)=alnx, ln(x^3)=3lnx,# and we get

#3(3)intxlnxdx=9intxlnxdx# factoring all constants outside.

Now, we'll make the following selections for Integration by Parts:

#u=lnx#

#du=x^-1dx#

#dv=xdx#

#v=intxdx=1/2x^2#

So, applying the formula, we get

#uv-intvdu=(x^2lnx)/2-1/2intx^-1x^2dx#

#=(x^2lnx)/2-1/2intxdx=(x^2lnx)/2-1/4x^2+C#

Recall that we need to multiply through by the #9# we factored out:

#9[(x^2lnx)/2-1/4x^2+C]=(9x^2lnx)/2-9/4x^2+C#

Thus,

#int3xln(x^3)dx=(9x^2lnx)/2-9/4x^2+C#

Elijah Abram · Answer

#int 3xlnx^3dx = (9x^2)/4 (2lnx-1)+C#

Using the properties of logarithms:

#ln x^3 = 3lnx #

so:

#int 3xlnx^3dx = 9 int xlnxdx#

integrate now by parts:

#int 3xlnx^3dx = 9 int lnx d(x^2/2)#

#int 3xlnx^3dx = (9x^2lnx)/2 - 9/2 int x^2 d(lnx)#

#int 3xlnx^3dx = (9x^2lnx)/2 - 9/2 int x^2 dx/x#

#int 3xlnx^3dx = (9x^2lnx)/2 - 9/2 int x dx#

#int 3xlnx^3dx = (9x^2lnx)/2 - (9x^2)/4+C#

#int 3xlnx^3dx = (9x^2)/4 (2lnx-1)+C#

Axel Alvarez · Answer

undefined To integrate ∫3xln(x^3) dx using integration by parts, let u = ln(x^3) and dv = 3x dx. Then differentiate u to find du, and integrate dv to find v. Afterward, apply the integration by parts formula: ∫udv = uv - ∫vdu. Finally, plug in the values to obtain the result.

HIX Tutor · Answer

undefined To integrate $ \int 3x \ln(x^3) \, dx $ using integration by parts, follow these steps:

1. **Simplify the integral**: 
   $$
   \ln(x^3) = 3 \ln x
   $$
   So, the integral becomes:
   $$
   \int 3x \ln(x^3) \, dx = \int 9x \ln x \, dx
   $$

2. **Choose $ u $ and $ dv $**:  
   Let:
   - $ u = \ln x $ (so $ du = \frac{1}{x} \, dx $)
   - $ dv = 9x \, dx $ (so $ v = \frac{9}{2} x^2 $)

3. **Apply integration by parts formula**:
   The formula is:
   $$
   \int u \, dv = uv - \int v \, du
   $$
   Now plug in $ u $, $ v $, $ du $, and $ dv $:
   $$
   \int 9x \ln x \, dx = \frac{9}{2} x^2 \ln x - \int \frac{9}{2} x^2 \cdot \frac{1}{x} \, dx
   $$

4. **Simplify the new integral**:
   $$
   \frac{9}{2} x^2 \cdot \frac{1}{x} = \frac{9}{2} x
   $$
   So:
   $$
   \int 9x \ln x \, dx = \frac{9}{2} x^2 \ln x - \int \frac{9}{2} x \, dx
   $$

5. **Integrate the remaining integral**:
   $$
   \int \frac{9}{2} x \, dx = \frac{9}{2} \cdot \frac{x^2}{2} = \frac{9}{4} x^2
   $$

6. **Combine results**:
   Now put it all together:
   $$
   \int 9x \ln x \, dx = \frac{9}{2} x^2 \ln x - \frac{9}{4} x^2 + C
   $$
   where $ C $ is the constant of integration.

So, the final answer is:
$$
\int 3x \ln(x^3) \, dx = \frac{9}{2} x^2 \ln x - \frac{9}{4} x^2 + C
$$