How do you integrate #int x^3lnabsx# from 1 to 2 by integration by parts method?

Question

Christopher Allers · Accepted Answer

#= 4 ln 2 - 15/16 # #int_1^2 x^3lnabsx \ dx# #= int_1^2 x^3 ln x \ dx#, as #x >0# #= int_1^2 d/dx(x^4/4) ln x \ dx# #= [ x^4/4 ln x ]_1^2 - int_1^2 x^4/4 d/dx( ln x )\ dx# #= [ x^4/4 ln x ]_1^2 - int_1^2 x^3/4 \ dx# #= [ x^4/4 ln x - x^4/16 ]_1^2# #= [ 16/4 ln 2 - 16/16 ] - [ 1/4 (0) - 1/16 ]# #= 4 ln 2 - 15/16 #

Charles Anctil · Answer

undefined To integrate $ \int x^3 \ln| x | \, dx $ from 1 to 2 using integration by parts:

1. Let $ u = \ln| x | $ and $ dv = x^3 \, dx $.
2. Find $ du $ and $ v $.
3. Apply integration by parts formula $ \int u \, dv = uv - \int v \, du $.
4. Calculate the definite integral from 1 to 2.

Here are the steps:

1. Let $ u = \ln| x | $ and $ dv = x^3 \, dx $.
   Then, $ du = \frac{1}{x} \, dx $ and $ v = \frac{x^4}{4} $.

2. Apply integration by parts formula:
   $$ \int x^3 \ln| x | \, dx = \frac{x^4 \ln| x |}{4} - \int \frac{x^4}{4} \cdot \frac{1}{x} \, dx $$

3. Simplify the integral:
   $$ = \frac{x^4 \ln| x |}{4} - \frac{1}{4} \int x^3 \, dx $$

4. Integrate the remaining integral:
   $$ = \frac{x^4 \ln| x |}{4} - \frac{1}{4} \cdot \frac{x^4}{4} \bigg|_1^2 $$

5. Evaluate the definite integral:
   $$ = \left( \frac{2^4 \ln| 2 |}{4} - \frac{1}{4} \cdot \frac{2^4}{4} \right) - \left( \frac{1^4 \ln| 1 |}{4} - \frac{1}{4} \cdot \frac{1^4}{4} \right) $$

6. Simplify and compute:
   $$ = \left( \frac{16 \ln| 2 |}{4} - \frac{16}{4} \right) - \left( \frac{0}{4} - \frac{1}{16} \right) $$
   $$ = \left( 4 \ln| 2 | - 4 \right) - \left( - \frac{1}{16} \right) $$
   $$ = 4 \ln 2 - 4 + \frac{1}{16} $$

So, the integral of $ \int_{1}^{2} x^3 \ln| x | \, dx $ is $ 4 \ln 2 - 4 + \frac{1}{16} $.