Evaluate the integrals in Exercises 31-50 . Some integrals do not require integration by parts.

step 1 Problem 33 says that we have to integrate x, lnx whole square dx. Let's substitute lnx as t. So

Evaluate the integrals in Exercises 31-50 . Some integrals...

Key Concepts

Selection of u and dv

The strategy of choosing which part of the integrand to assign as u and which as dv is essential to the effective use of integration by parts. A good rule of thumb is to choose u as the function that becomes simpler upon differentiation (such as logarithmic functions) and dv as the part that can be easily integrated (such as polynomial functions). This careful selection reduces the overall complexity of the integral and often leads to a solution in fewer steps.

Integration by Parts

Integration by parts is a fundamental technique used to integrate products of functions and is based on the product rule for differentiation. Its formula ? u dv = u v ? ? v du allows one to break down a complex integral into simpler parts by appropriately choosing u and dv. This method is particularly useful when the integrand involves a logarithmic or algebraic factor, where differentiating one part simplifies it while integrating the other part does not complicate it further.

Differentiation of Logarithmic Functions

Logarithmic functions have derivatives that simplify the integration process when used in combination with other functions. Specifically, the derivative of ln x is 1/x, which often cancels or reduces the complexity of the integrand. Understanding how to differentiate logarithmic functions is crucial when they are selected as the u component in integration by parts, as it typically leads to a more manageable integral.

Transcript

00:01 Problem 33 says that we have to integrate x, lnx whole square dx.

00:11 Let's substitute lnx as t.

00:14 So this means that x is e -raised to the power t, and this means that dx will be e -raised to the power t d t.

00:22 So if this was integral i, so now the integral i becomes integral of x is replaced by erase to the power t, lnx is t and dx is e raised to the power t and dt.

00:35 So this becomes integral of t square e raised to the power 2t dt.

00:41 Now this is the first term, this is the second term.

00:43 We have to use integration by parts.

00:46 So the first term remains as it is we integrate the second term which is e raised to the power 2t over 2 minus integral of differentiation of t square is 2t.

00:55 Integral of e raised to the power 2 t is e raised to the power 2t over 2 and this one is again integrated.

01:00 So the value of i comes out as t squared e raised to the power 2t over 2 minus once again integration by parts so first term and second term so the first term remains as it is integral of second term will be a yes to the power 2 t over 2 minus integral integral of differentiation of t will be 1 integral of e raised to the power 2 t will be a raised to the power 2 t over 2 and this one should again be integrated with respect 2 so i will read at the value of i i is t square e raised to the power 2 is over 2 t over 2 minus t e raised to the power 2 t over 2 minus this will be 1 over 2 over here and this will be integral of e raise to the power 2 t t so we integrate this further and let's open up the brackets as well so we have minus t over 2 e raised to the power 2 t here and we have 1 over 2 here and we have 1 over 2 here and this has to be integrated...

Please give Ace some feedback