# int_(-1)^1x^2/(sqrt(x^2+1)+x+1)dx = ?

## Find ${\int}_{- 1}^{1} {x}^{2} / \left(\sqrt{{x}^{2} + 1} + x + 1\right) \mathrm{dx}$

Apr 3, 2018

${\int}_{- 1}^{1} {x}^{2} / \left(\sqrt{{x}^{2} + 1} + x + 1\right) \cdot \mathrm{dx} = \frac{1}{3}$

#### Explanation:

${\int}_{- 1}^{1} {x}^{2} / \left(\sqrt{{x}^{2} + 1} + x + 1\right) \cdot \mathrm{dx}$

=${\int}_{- 1}^{1} \frac{{x}^{2} \cdot \left(x + 1 - \sqrt{{x}^{2} + 1}\right) \cdot \mathrm{dx}}{{\left(x + 1\right)}^{2} - \left({x}^{2} + 1\right)}$

=${\int}_{- 1}^{1} \frac{{x}^{2} \cdot \left(x + 1 - \sqrt{{x}^{2} + 1}\right) \cdot \mathrm{dx}}{2 x}$

=$\frac{1}{2} {\int}_{- 1}^{1} x \cdot \left(x + 1 - \sqrt{{x}^{2} + 1}\right) \cdot \mathrm{dx}$

=$\frac{1}{2} {\int}_{- 1}^{1} \left({x}^{2} + x - x \sqrt{{x}^{2} + 1}\right) \cdot \mathrm{dx}$

=${\left[\frac{1}{6} {x}^{3} + \frac{1}{4} {x}^{2} - \frac{1}{6} {\left({x}^{2} + 1\right)}^{\frac{3}{2}}\right]}_{- 1}^{1}$

=$\frac{1}{3}$

Apr 3, 2018

${\int}_{- 1}^{1} {x}^{2} / \left(\sqrt{{x}^{2} + 1} + x + 1\right) \cdot \mathrm{dx} = \frac{1}{3}$

#### Explanation:

${\int}_{- 1}^{1} {x}^{2} / \left(\sqrt{{x}^{2} + 1} + x + 1\right) \cdot \mathrm{dx}$

After using $x = \tan y$ and $\mathrm{dx} = {\left(\sec y\right)}^{2} \cdot \mathrm{dy}$ transforms, this integral became

${\int}_{- \frac{\pi}{4}}^{\frac{\pi}{4}} {\left(\tan y\right)}^{2} / \left(\sec y + \tan y + 1\right) \cdot {\left(\sec y\right)}^{2} \cdot \mathrm{dy}$

=${\int}_{- \frac{\pi}{4}}^{\frac{\pi}{4}} \frac{{\left(\tan y\right)}^{2} \cdot \left(\tan y + 1 - \sec y\right) \cdot {\left(\sec y\right)}^{2} \cdot \mathrm{dy}}{{\left(\tan y + 1\right)}^{2} - {\left(\sec y\right)}^{2}}$

=${\int}_{- \frac{\pi}{4}}^{\frac{\pi}{4}} \frac{{\left(\tan y\right)}^{2} \cdot \left(\tan y + 1 - \sec y\right) \cdot {\left(\sec y\right)}^{2} \cdot \mathrm{dy}}{{\left(\tan y\right)}^{2} + 1 + 2 \tan y - {\left(\sec y\right)}^{2}}$

=${\int}_{- \frac{\pi}{4}}^{\frac{\pi}{4}} \frac{{\left(\tan y\right)}^{2} \cdot \left(\tan y + 1 - \sec y\right) \cdot {\left(\sec y\right)}^{2} \cdot \mathrm{dy}}{{\left(\sec y\right)}^{2} + 2 \tan y - {\left(\sec y\right)}^{2}}$

=${\int}_{- \frac{\pi}{4}}^{\frac{\pi}{4}} \frac{{\left(\tan y\right)}^{2} \cdot \left(\tan y + 1 - \sec y\right) \cdot {\left(\sec y\right)}^{2} \cdot \mathrm{dy}}{2 \tan y}$

=$\frac{1}{2} {\int}_{- \frac{\pi}{4}}^{\frac{\pi}{4}} \tan y \cdot \left(\tan y + 1 - \sec y\right) \cdot {\left(\sec y\right)}^{2} \cdot \mathrm{dy}$

=$\frac{1}{2} {\int}_{- \frac{\pi}{4}}^{\frac{\pi}{4}} {\left(\tan y\right)}^{2} \cdot {\left(\sec y\right)}^{2} \cdot \mathrm{dy}$+$\frac{1}{2} {\int}_{- \frac{\pi}{4}}^{\frac{\pi}{4}} \left(\tan y\right) \cdot {\left(\sec y\right)}^{2} \cdot \mathrm{dy}$-$\frac{1}{2} {\int}_{- \frac{\pi}{4}}^{\frac{\pi}{4}} {\left(\sec y\right)}^{3} \cdot \tan y \cdot \mathrm{dy}$

=${\left[\frac{1}{6} {\left(\tan y\right)}^{3} + \frac{1}{4} {\left(\tan y\right)}^{2} - \frac{1}{6} {\left(\sec y\right)}^{3}\right]}_{- \frac{\pi}{4}}^{\frac{\pi}{4}}$

=$\frac{1}{3}$