# How do you find the definite integral of (x dx) / [(x^2 +1)(ln (x^2 +1))] from [1, 2]?

May 10, 2018

$I = \frac{1}{2} \ln | \ln \frac{5}{\ln} 2 |$

$\mathmr{and} I \approx 0.42$

#### Explanation:

We have,

$I = {\int}_{1}^{2} \frac{x}{\left({x}^{2} + 1\right) \left(\ln \left({x}^{2} + 1\right)\right)} \mathrm{dx}$

Subst.

$\ln \left({x}^{2} + 1\right) = u \implies {x}^{2} + 1 = {e}^{u} \implies 2 x \mathrm{dx} = {e}^{u} \mathrm{du}$

$\implies x \mathrm{dx} = \frac{1}{2} {e}^{u} \mathrm{du}$

$x = 1 \implies u = \ln \left({1}^{2} + 1\right) = \ln 2 = a , \to \left(s a y\right)$

$x = 2 \implies u = \ln \left({2}^{2} + 1\right) = \ln 5 = b , \to \left(s a y\right)$

So,

$I = {\int}_{a}^{b} \frac{\frac{1}{2} {e}^{u}}{{e}^{u} \left(u\right)} \mathrm{du}$

$I = \frac{1}{2} {\int}_{a}^{b} \frac{1}{u} \mathrm{du}$

$= \frac{1}{2} {\left[\ln u\right]}_{a}^{b}$

$= \frac{1}{2} \left[\ln b - \ln a\right]$

$= \frac{1}{2} \ln | \frac{b}{a} | , w h e r e , a = \ln 2 \mathmr{and} b = \ln 5$

$I = \frac{1}{2} \ln | \ln \frac{5}{\ln} 2 |$

$I \approx 0.42$

f(x)=x/((x^2+1)(ln(x^2+1)) is continuous in $\left[1 , 2\right]$

graph{x/((x^2+1)(ln(x^2+1))) [-4, 6, -1.7, 3.3]}

May 10, 2018

${\int}_{1}^{2} \frac{x \mathrm{dx}}{\left({x}^{2} + 1\right) \left(\ln \left({x}^{2} + 1\right)\right)} = \frac{1}{2} {\left[\ln \left(\ln \left({x}^{2} + 1\right)\right)\right]}_{1}^{2} = \frac{\ln \left(\ln \left(5\right)\right) - \ln \left(\ln \left(2\right)\right)}{2} = 0.421198958$

#### Explanation:

show below

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

suppose:

$u = {x}^{2} + 1$

$\mathrm{du} = 2 x \cdot \mathrm{dx}$

$\mathrm{dx} = \frac{\mathrm{du}}{2 x}$

$x = \sqrt{u - 1}$

${\int}_{1}^{2} \frac{\sqrt{u - 1} \cdot \frac{\mathrm{du}}{\sqrt{u - 1}}}{u \cdot \ln u}$

$\frac{1}{2} {\int}_{1}^{2} \frac{\mathrm{du}}{u \cdot \ln u}$

$\frac{1}{2} {\int}_{1}^{2} \frac{\frac{1}{u} \cdot \mathrm{du}}{\ln u} = \frac{1}{2} {\left[\ln \left(\ln \left(u\right)\right)\right]}_{1}^{2}$

$\frac{1}{2} {\left[\ln \left(\ln \left({x}^{2} + 1\right)\right)\right]}_{1}^{2} = \frac{\ln \left(\ln \left(5\right)\right) - \ln \left(\ln \left(2\right)\right)}{2} = 0.421198958$