The derivative of the logarithm ln x lnx is 1 x x1, but what is the antiderivative? This turns out to be a little trickier, and has to be done using a clever integration by parts. The logarithm is a basic function from which many other functions are built, so learning to integrate it substantially broadens the kinds of integrals we can tackle. To evaluate the integral ∫ logxdx, we can use integration by parts. Let's go through the solution step by step. Step 1: Set up the integral Let I = ∫ logxdx. Step 2: Choose u and dv We will use integration by parts, which states: ∫ udv= uv−∫ vdu Here, we choose: - u= logx (which we will differentiate) - dv= dx (which we will integrate ) Step 3: Differentiate u and integrate dv Now we need to find du and v: - Differentiate u: du = 1 x dx - Integrate dv: v= x Step 4: Apply integration ... Transcript Ex 7.6, 4 log 1 log 1 log = 1 ( log ) = log 1 1 ( ( log )/ 1 . ) = log ( ^2/2) 1 1/ . ^2/2. = ^2/2 log 1/2 1 . = ^2/2 log 1/2 . ^2/2 + = ^ / ^ / + " " Show More To solve the integral ∫ logx x dx, we can use the substitution method. Here’s a step-by-step solution: Step 1: Substitution Let t= logx. Then, we differentiate both sides to find dx: dt = 1 x dx ⇒ dx = xdt Since x = et (from t= logx), we can substitute dx in terms of t: dx = etdt Step 2: Rewrite the Integral Now, substituting t and dx into the integral : ∫ logx x dx = ∫ t⋅ 1 x ⋅dx= ∫ t⋅dt Step 3: Integrate Now we can integrate t: ∫ tdt= t2 2 +C Step 4: Back Substitute Now ...
