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,

$\frac{d}{dx}\int_2^{x^2}ln(t^2+1)dt$

Required

The value of the differentiation.

Here, first integrating the function with respect to t.

By using the multiplicative property of integration,

$first\text{ function}\int second\text{ function-}\int(\frac{d}{dx}first\text{ function}\int second\text{ function\rparen}$

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

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

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.

$\begin{gathered} \frac{d}{dx}\int_2^{x^2}ln(t^2+1)dt=\frac{d}{dx}(x^2ln(x^4+1)-2x^2-2tan^{-1}x^2-2ln(5)+4-2tan^{-1}(2)) \\ =x^2\times\frac{1}{x^4+1}\times4x^3+2xln(x^4+1)-4x+\frac{4x}{x^4+1}-0+0-0 \\ =\frac{4x^5}{x^4+1}+\frac{4x}{x^4+1}-4x+(2x)ln(x^4+1) \\ =\frac{4x^5+4x-4x^5-4x}{x^4+1}+(2x)ln(x^4+1) \\ =(2x)ln(x^4+1) \end{gathered}$

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