Integral of xlnx

The integral of xlnx is equal to (x²/2) lnx​ - x²/4​ + C, where C is the integration constant. We can evaluate this integral using the method of integration by parts. We use the method of integration by parts to find the integral of the product of two functions. To find the integral of xlnx, we can consider xlnx as the product of two functions x and lnx. Mathematically, we can write the formula for the integral of xlnx as ∫xlnx dx = (x²/2) lnx​ - x²/4​ + C. We know that integration is the reverse process of differentiation and using the formula of the integration by parts, we can find the integral of xlnx.

Further in this article, we will find the integral of xlnx and derive its formula. We will also find the integral of functions involving xlnx and solve a few examples using the integral of xlnx for a better understanding of the concept.

The integral of xlnx is equal to (x²/2) lnx​ - x²/4​ + C, where C is the constant of integration. We can calculate the integration of xlnx using integration by parts which is one of the most commonly used and important methods of integration. We can write the integral of xlnx with symbols as ∫xlnx dx = (x²/2) lnx​ - x²/4​ + C, where ∫ is the symbol of integration, dx indicates that the integration of xlnx is with respect to the variable x and C is the integration constant. As we know that integration is the reverse process of differentiation, therefore we can say that the integral of xlnx is nothing but the antiderivative of xlnx. In the next section, let us go through its formula:

The formula for the integral of xlnx is given by, ∫xlnx dx = (x²/2) lnx​ - x²/4​ + C with C as the constant of integration. This formula can be derived using the integration by parts method of integration. The image given below shows the formula for the integral of xlnx:

Now that we know that the integral of xlnx is equal to x² lnx​/2 - x²/4​ + C, we will prove it using the method of integration by parts formula. For the integral of f(x) g(x), the formula is given by, ∫f(x) g(x) dx = f(x) ∫g(x) dx - ∫[f'(x) ∫g(x) dx] dx. For the integral of xlnx we will choose the first function and second function using the sequence ILATE (I - Inverse, L - Logarithmic, A - Algebraic, T - Trigonometric, E - Exponential), therefore we have f(x) = lnx and g(x) = x. So, we will need the integral of x and derivative of lnx formulas. Therefore, we can find the integral of xlnx as

∫xlnx dx = ln(x) ∫x dx - ∫[(lnx)' ∫x dx] dx

= lnx × x²/2 - ∫[(1/x) × x²/2] dx

= (x²/2) lnx​ - (1/2) ∫x dx​

= (x²/2) lnx​ - (1/2)(x²/2) + C

= (x²/2) lnx​ - x²/4​ + C

Hence, we have derived the formula for the integral of xlnx using the method of integration by parts.

To find the integral of x ln x by √(x² - 1), we will use the formula of the integration by parts as above. The formula for the integration by parts is given by, ∫f(x) g(x) dx = f(x) ∫g(x) dx - ∫[f'(x) ∫g(x) dx] dx. We assume f(x) = ln x and g(x) = x/√(x² - 1). Before evaluating the integral of x lnx / √(x² - 1), let us calculate the integral of x/√(x² - 1). To find this, assume u² = (x² - 1). Now, differentiating both sides, we have 2u du = 2x dx which implies xdx = udu. So, we have

∫[x/√(x² - 1)] dx = ∫(u/u) du

= ∫du

= u + K

= √(x² - 1) + K --- (1)

Next, we will calculate the integral of √(x² - 1)/x. To find this, assume v = √(x² - 1), then dv/dx = x/√(x² - 1) which implies dx = √(x² - 1) dv/x = v/√(v² + 1) dv and √(x² - 1)/x = v/√(v² + 1). Therefore, we have

∫[√(x² - 1)/x] dx = ∫ [v/√(v² + 1)] [v/√(v² + 1) dv]

= ∫[v²/(v² + 1)] dv

= ∫[(v² + 1 - 1)/(v² + 1)] dv

= ∫(v² + 1)/(v² + 1) dv - ∫1/(v² + 1) dv

= ∫dv - ∫1/(v² + 1) dv

= v - arctan(v) + M, where M is the integration constant.

= √(x² - 1) - arctan(√(x² - 1)) --- (2)

Now, we will calculate the integral of xlnx as

∫x lnx/√(x² - 1) dx = ∫[x/√(x² - 1)] lnx dx

= lnx ∫[x/√(x² - 1)] dx - ∫[(lnx)' ∫[x/√(x² - 1)] dx] dx

= ln x × (√(x² - 1)) - ∫[(1/x) √(x² - 1)] dx --- [From (1)]

= √(x² - 1) ln x - [√(x² - 1) - arctan(√(x² - 1))] --- [From (2)]

= √(x² - 1) ln x - √(x² - 1) + arctan(√(x² - 1)) + C, where C is the integration of constant.

Hence, the integral of x lnx / √(x² - 1) is equal to √(x² - 1) ln x - √(x² - 1) + arctan(√(x² - 1)) + C.

Important Notes on Integral of xlnx

• The integral of xlnx is equal to (x²/2) lnx​ - x²/4​ + C.
• We can find the integration of xlnx using the integration by parts method.

☛ Related Topics:

• Integral of 2sinx
• Integration of root x
• Integral of secx tanx

What is Integral of xlnx in Calculus?

The integral of xlnx is equal to (x²/2) lnx​ - x²/4​ + C, where C is the integration constant. We can evaluate this integral using the method of integration by parts.

What is the Formula for the Integral of xlnx?

The formula for the integral of xlnx is given by, ∫xlnx dx = (x²/2) lnx​ - x²/4​ + C with C as the constant of integration.

What is the Definite Integral of xlnx From 0 to 1?

The definite integral of xlnx from 0 to 1 is equal to -1/4. We can evaluate this integral by substituting the limits 0 and 1 into the formula of the integral of xlnx given by, (x²/2) lnx​ - x²/4​ + C.

How to Find the Integral of xlnx?

We can calculate the integral of xlnx using the method of integration by parts. We can assume f(x) = lnx and g(x) = x and substitute into the formula ∫f(x) g(x) dx = f(x) ∫g(x) dx - ∫[f'(x) ∫g(x) dx] dx to find the integral of xlnx.

How to Calculate the Integral of xlnx - 1?

We can calculate the integral of xlnx - 1 using the formula ∫[f(x) - g(x)] dx = ∫f(x) dx - ∫g(x) dx. We can write the integral as ∫(xlnx - 1) dx = ∫xlnx dx - ∫dx = (x²/2) lnx​ - x²/4​ - x + C.