Question

Combining derivatives and integrals: Simplify each of the following as much as possible:

${\int }_{-2}^x\frac{d}{dx}\left(\ln \right(x^2+1\left)\right)dx$

Short answer

The solution is${\int }_{-2}^x\frac{d}{dx}\left(\ln \right(x^2+1\left)\right)dx=\left[\frac{\left|x^2+1\right|}{5}\right].$

Step-by-step solution

Step 1. Given information

Simplify:${\int }_{-2}^x\frac{d}{dx}\left(\ln \right(x^2+1\left)\right)dx$

Step 2. Calculation

The given equation is:

${\int }_{-2}^x\frac{d}{dx}\left(\ln \right(x^2+1\left)\right)dx$

$\frac{d}{dx}\ln x=\frac{1}{x}$& Chain rule

$$\begin{gathered} {\int }_{-2}^x\frac{d}{dx}\left(\ln \right(x^2+1\left)\right)dx={\int }_{-2}^x\left(\frac{1}{(x^2+1)dx}\frac{d}{dx}\right(x^2+1\left)\right)dx \\ {\int }_{-2}^x\frac{d}{dx}\left(\ln \right(x^2+1\left)\right)dx={\int }_{-2}^x\left(\frac{2x}{(x^2+1)}\right)dx \\ Let(x^2+1)=t \end{gathered}$$

Differentiate both sides with respect to 't'

$$\begin{gathered} \frac{d}{dt}(x^2+1)=\frac{dt}{dt} \\ 2xdx=dt \end{gathered}$$

Also when

$x=-2thent=5\&whenx=xthent=(x^2+1)$

$$\begin{gathered} {\int }_{-2}^x\frac{d}{dx}\left(\ln \right(x^2+1\left)\right)dx={\int }_5^{x^2+1}\left(\frac{1}{t}\right)dt \\ U\sin g\int \frac{1}{x}dx=\ln \left|x\right|+C \\ {\int }_{-2}^x\frac{d}{dx}\left(\ln \right(x^2+1\left)\right)dx={\left[\ln \left|t\right|\right]}_5^{x^2+1}{\int }_{-2}^x\frac{d}{dx}\left(\ln \right(x^2+1\left)\right)dx=[\ln \left|x^2+1\right|-\ln 5] \\ {\int }_{-2}^x\frac{d}{dx}\left(\ln \right(x^2+1\left)\right)dx=\left[\ln \frac{\left|x^2+1\right|}{5}\right] \end{gathered}$$