Learn Evaluate The Integral Logx 1 Logx Dx Xx 1

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# Evaluate the integral log(x+1) - log x dx/x(x+1).

To evaluate the integral value of log (x+1) - log x dx/x (x+1), we will use substutitution method.

## Answer: ∫ log (x + 1) - log x dx/x (x + 1) = -1/2(log (x + 1))^{2} - 1/2 (log x)^{2} - log (x + 1) log x + C

Here is a detailed explanation.

**Explanation:**

Let log (x+1) - log x = t.

On differentiating the above equation with respect to 'x', we get

1/(x+1) - 1/x = dt/dx

d/dx log x = 1/x

On solving this we'll get,

-1/x (x+1) = dt/dx

dx = -x (x+1)dt

On substituting the value t and dt in the equation,we get

∫ t/x (x+1) - x (x+1) dt

On solving this we get

∫ -t dt = -t^{2}/2 + C

On substituting the values of 't' in the equation, we get

-1/2 [log (x+1) - log x]^{2 }+ C

-1/2(log (x+1))^{2} - 1/2 (log x)^{2} - log (x+1) log x + C

### Thus, the integral value of log (x+1) - log x dx/x (x+1) is -1/2(log (x+1))^{2} - 1/2 (log x)^{2} - log (x+1) log x + C

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