Question

Find the derivative with respect to x of the integral from 2 to x squared of the quantity the natural lot of the quantity t squared plus 1, dt

Answer 1

Given,

The expression is,

`(d)/(dx)\int_2^(x^2)ln(t^2+1)dt`

Required

The value of the differentiation.

Here, first integrating the function with respect to t.

By using the multiplicative property of integration,

`first\text{ function}\int second\text{ function-}\int((d)/(dx)first\text{ function}\int second\text{ function\rparen}`

Here, first function is ln(t^2+1) and second function is 1.

`\begin{gathered} \int ln(t^2+1)dt=ln(t^2+1)\int dt\text{-}\int((d)/(dx)ln(t^2+1)\int1dt\text{\rparen} \\ =ln(t^2+1)* t\text{-}\int(2t^2)/(t^2+1)dt \\ =ln(t^2+1)* t\text{-2}\int((t^2+1)/(t^2+1)-(1)/(t^2+1))dt \\ =ln(t^2+1)* t\text{-2}\int(1-(1)/(t^2+1))dt \\ =ln(t^2+1)* t\text{-2\lparen t-tan}^(-1)t) \\ =tln(t^2+1)\text{-2t+2tan}^(-1)t) \end{gathered}`

Substituting the limits over the integrated value.

`\begin{gathered} \int_2^(x2)ln(t^2+1)dt=(tln(t^2+1)\text{-2t-2tan}^(-1)t)_2^(x^2) \\ =x^2ln(x^4+1)-2x^2+2tan^(-1)(x^2)-2ln(5)+4-2tan^(-1)(2) \end{gathered}`

Now, differentiating the expression obtained by integrating the given function.

`\begin{gathered} (d)/(dx)\int_2^(x^2)ln(t^2+1)dt=(d)/(dx)(x^2ln(x^4+1)-2x^2-2tan^(-1)x^2-2ln(5)+4-2tan^(-1)(2)) \\ =x^2*(1)/(x^4+1)*4x^3+2xln(x^4+1)-4x+(4x)/(x^4+1)-0+0-0 \\ =(4x^5)/(x^4+1)+(4x)/(x^4+1)-4x+(2x)ln(x^4+1) \\ =(4x^5+4x-4x^5-4x)/(x^4+1)+(2x)ln(x^4+1) \\ =(2x)ln(x^4+1) \end{gathered}`

Hence, the derivative of the function is (2x) ln (x^4+1).