What technique can be used to solve the integral of x^3 ln(x) dx?

What technique can be used to solve the integral of x^3 ln(x) dx?

• A. Substitution
• B. Integration by parts
• C. Partial fractions
• D. Trigonometric substitution

Answer: (B)

To solve the integral of \( x^3 \ln(x) \, dx \), integration by parts is the most effective technique to use. This approach is particularly useful here since the integral involves a product of a polynomial function, \( x^3 \), and a logarithmic function, \( \ln(x) \). In integration by parts, the formula used is: \[ \int u \, dv = uv - \int v \, du \] For the integral in question, you can choose \( u = \ln(x) \) and \( dv = x^3 \, dx \). This selection is beneficial because it simplifies the logarithmic function through differentiation (where \( du = \frac{1}{x} \, dx \)), while integrating the polynomial \( x^3 \) leads to a simpler polynomial \( v = \frac{x^4}{4} \). After applying the formula, the first term \( uv \) will yield \( \frac{x^4}{4} \ln(x) \) while the remaining integral \( \int v \, du \), which becomes \( \int \frac{x^4}{4} \cdot \frac{1}{x} \

To solve the integral of ( x^3 \ln(x) , dx ), integration by parts is the most effective technique to use. This approach is particularly useful here since the integral involves a product of a polynomial function, ( x^3 ), and a logarithmic function, ( \ln(x) ).

In integration by parts, the formula used is:

[

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

]

For the integral in question, you can choose ( u = \ln(x) ) and ( dv = x^3 , dx ). This selection is beneficial because it simplifies the logarithmic function through differentiation (where ( du = \frac{1}{x} , dx )), while integrating the polynomial ( x^3 ) leads to a simpler polynomial ( v = \frac{x^4}{4} ).

After applying the formula, the first term ( uv ) will yield ( \frac{x^4}{4} \ln(x) ) while the remaining integral ( \int v , du ), which becomes ( \int \frac{x^4}{4} \cdot \frac{1}{x} \