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plzz integrate 1/loogx-1/(logx)^2
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plzz integrate 1/loogx-1/(logx)^2
Integration of 1/log(x) - 1/(log(x))^2:

To integrate the given expression, 1/log(x) - 1/(log(x))^2, we can use the method of substitution. Let's break down the integration step by step:

Step 1: Rewrite the expression:
First, let's rewrite the expression in a more convenient form by taking the common denominator.

1/log(x) - 1/(log(x))^2 = (log(x) - 1)/(log(x))^2

Step 2: Substitute a variable:
Let's substitute a variable to simplify the expression. We can let u = log(x), which implies du/dx = 1/x.

Step 3: Rewrite the expression in terms of u:
By substituting u = log(x), the expression becomes:

(log(x) - 1)/(log(x))^2 = (u - 1)/u^2

Step 4: Differentiate u with respect to x:
To find du/dx, we differentiate u = log(x) with respect to x using the chain rule:

du/dx = (1/x)

Step 5: Rewrite the expression in terms of u and du/dx:
By substituting the values of u and du/dx, the expression becomes:

(u - 1)/u^2 = (u - 1)/(u^2 * (1/x))

Step 6: Simplify the expression:
To simplify further, we can rewrite the expression as:

(u - 1)/(u^2 * (1/x)) = (x * (u - 1))/u^2

Step 7: Integrate the expression:
Now we can integrate the simplified expression:

∫[(x * (u - 1))/u^2] dx = ∫(x(u - 1))/u^2 dx

Step 8: Simplify the integral:
To simplify the integral, we can rewrite it as:

∫(x/u^2 - x/u) dx

Step 9: Integrate each term separately:
Now we can integrate each term separately:

∫(x/u^2 - x/u) dx = ∫(x/u^2) dx - ∫(x/u) dx

Step 10: Apply the power rule of integration:
Using the power rule of integration, the integral of x/u^2 can be written as:

∫(x/u^2) dx = ∫x * u^-2 dx = x * (u^-1)/(-1) + C1

Similarly, the integral of x/u can be written as:

∫(x/u) dx = ∫x * u^-1 dx = x * (u^0)/(0) + C2

Step 11: Combine the integrals:
Combining the integrals, we have:

x * (u^-1)/(-1) + C1 - x * (u^0)/(0) + C2

Since (u^0)/(0) is undefined, we can simplify the expression as:

-x/u + C1 + C2

Step 12: Simplify the expression:
Community Answer
plzz integrate 1/loogx-1/(logx)^2
I = {1/logx -1/(logx)2 }dxmultiply numerator & denominator by xI = x{1/logx-1/(logx)2}dx /xput logx = t dx/x = dt & x=etnow I = et{1/t - 1/t2}dt I = et{ f(t) + f1(t) }we have a direct formula for ex{f(x) + f1(x)}dx = exf(x) so I = et f(t) =et/t = (x / logx) lim 0 to e2 =0 - e2/2
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plzz integrate 1/loogx-1/(logx)^2
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