54. What is the solution of ∫ 𝑥2 ln 𝑥𝑑𝑥 ? Given C is an arbitrary constant.
Evaluate ∫ x² ln x dx Using Integration by Parts
Understanding the Given Integration Problem
The given problem asks us to evaluate the indefinite integral:
∫ x2 ln x dx
This integral contains the product of two different types of functions. The first function, x², is an algebraic function, while the second function, ln x, is a logarithmic function. There is no direct standard formula that immediately integrates the product x² ln x. Therefore, the most suitable method is integration by parts.
Integration by parts is especially useful when an integral contains a logarithmic function multiplied by an algebraic function. The main objective is to differentiate the logarithmic part because the derivative of ln x becomes the much simpler expression 1/x.
Using the Integration by Parts Formula
The standard formula for integration by parts is:
∫ u dv = uv − ∫ v du
To use this formula correctly, we must divide the integrand into two parts. One part is selected as u, which will be differentiated, while the remaining part is selected as dv, which will be integrated.
For the given integral:
∫ x2 ln x dx
we choose:
u = ln x
and:
dv = x2 dx
This choice simplifies the integral because differentiating ln x gives 1/x, while integrating x² is straightforward.
Finding du and v
Differentiating u = ln x
We have selected:
u = ln x
Differentiating both sides with respect to x gives:
du/dx = 1/x
Therefore:
du = (1/x) dx
Integrating dv = x² dx
We have:
dv = x2 dx
Integrating both sides:
v = ∫ x2 dx
Using the power rule of integration:
∫ xn dx = xn+1/(n + 1)
we obtain:
v = x3/3
Thus, the required quantities are:
u = ln x
du = (1/x) dx
dv = x2 dx
v = x3/3
Applying the Integration by Parts Formula
Now use the formula:
∫ u dv = uv − ∫ v du
Substituting the values of u, v, and du:
∫ x2 ln x dx = (ln x)(x3/3) − ∫ (x3/3)(1/x) dx
Therefore:
∫ x2 ln x dx = (x3 ln x)/3 − (1/3)∫ x3/x dx
Since:
x3/x = x2
the expression becomes:
∫ x2 ln x dx = (x3 ln x)/3 − (1/3)∫ x2 dx
Evaluating the Remaining Integral
The remaining integral is:
∫ x2 dx
Using the power rule:
∫ x2 dx = x3/3
Therefore:
∫ x2 ln x dx = (x3 ln x)/3 − (1/3)(x3/3) + C
Simplifying the second term:
(1/3)(x3/3) = x3/9
Hence:
∫ x2 ln x dx = (x3 ln x)/3 − x3/9 + C
Therefore, the required solution is:
(x3/3) ln x − x3/9 + C
Writing the Answer in an Alternative Equivalent Form
The result:
(x3 ln x)/3 − x3/9 + C
can also be written by taking x³/9 as a common factor.
Since:
(x3 ln x)/3 = (x3/9)(3 ln x)
we obtain:
∫ x2 ln x dx = (x3/9)(3 ln x − 1) + C
Thus, both of the following forms are equivalent:
(x3 ln x)/3 − x3/9 + C
and:
(x3/9)(3 ln x − 1) + C
Verification of the Answer by Differentiation
An indefinite integral can be verified by differentiating the obtained answer. Consider:
F(x) = (x3 ln x)/3 − x3/9 + C
Differentiate the first term using the product rule:
d/dx [(x3/3) ln x]
= (x3/3)(1/x) + ln x · d/dx(x3/3)
Therefore:
= x2/3 + x2 ln x
Now differentiate the second term:
d/dx (−x3/9) = −3x2/9
Therefore:
d/dx (−x3/9) = −x2/3
Combining the derivatives:
x2/3 + x2 ln x − x2/3
The terms x²/3 and −x²/3 cancel, leaving:
x2 ln x
This is exactly the original integrand. Therefore, the obtained antiderivative is correct.
Why Integration by Parts Is the Best Method
The expression x² ln x is a product of an algebraic function and a logarithmic function. Differentiating ln x simplifies it to 1/x, while integrating x² produces x³/3. Their combination reduces the original integral to the simple power integral ∫x² dx.
This makes integration by parts the natural and efficient method for solving the problem. The logarithmic function should be selected as u because its derivative is simpler than the original function.
Final Answer
The given integral is:
∫ x2 ln x dx
Using integration by parts:
u = ln x
dv = x2 dx
Therefore:
du = (1/x) dx
v = x3/3
Hence:
∫ x2 ln x dx = (x3 ln x)/3 − x3/9 + C
Correct Answer: (x3 ln x)/3 − x3/9 + C
