# Question c5bb3

Dec 20, 2017

3) $\int {\left(\arctan x\right)}^{2} / \left(1 + {x}^{2}\right) \cdot \mathrm{dx}$

=${\left(\arctan x\right)}^{3} / 3 + C$

#### Explanation:

3) $\int {\left(\arctan x\right)}^{2} / \left(1 + {x}^{2}\right) \cdot \mathrm{dx}$

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

$\int \frac{{u}^{2} \cdot {\left(\sec u\right)}^{2} \cdot \mathrm{du}}{\sec u} ^ 2$

=${u}^{2} \cdot \mathrm{du}$

=${u}^{3} / 3 + C$

=${\left(\arctan x\right)}^{3} / 3 + C$

Dec 20, 2017

1) $\int \left({x}^{2} - 3 x\right) \cdot {\left(\cos x\right)}^{2} \cdot \mathrm{dx}$

=${x}^{3} / 6 - \frac{3 {x}^{2}}{4} + \frac{2 {x}^{2} - 6 x - 1}{8} \cdot \sin 2 x + \frac{2 x - 3}{8} \cdot \cos 2 x + C$

#### Explanation:

1) $\int \left({x}^{2} - 3 x\right) \cdot {\left(\cos x\right)}^{2} \cdot \mathrm{dx}$

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

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

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

$B = \int \left({x}^{2} - 3 x\right) \cdot \cos 2 x$

=$\left({x}^{2} - 3 x\right) \cdot \frac{1}{2} \sin 2 x - \left(2 x - 3\right) \left(- \frac{1}{4} \cos 2 x\right) + 2 \left(- \frac{1}{8} \sin 2 x\right)$

=$\frac{2 {x}^{2} - 6 x - 1}{4} \cdot \sin 2 x + \frac{2 x - 3}{4} \cdot \cos 2 x$

Thus,

$\int \left({x}^{2} - 3 x\right) \cdot {\left(\cos x\right)}^{2} \cdot \mathrm{dx}$

=$\frac{1}{2} \cdot A + \frac{1}{2} \cdot B$

=${x}^{3} / 6 - \frac{3 {x}^{2}}{4} + \frac{2 {x}^{2} - 6 x - 1}{8} \cdot \sin 2 x + \frac{2 x - 3}{8} \cdot \cos 2 x + C$

Dec 20, 2017

$\int \frac{{x}^{3} + 2}{{x}^{2} - x + 1} \cdot \mathrm{dx}$

=${x}^{2} / 2 + x + \frac{2 \sqrt{3}}{3} \arctan \left(\frac{2 x - 1}{\sqrt{3}}\right) + C$

#### Explanation:

2)# $\int \frac{{x}^{3} + 2}{{x}^{2} - x + 1} \cdot \mathrm{dx}$

=$\int \frac{{x}^{3} + 1}{{x}^{2} - x + 1} \cdot \mathrm{dx}$+$\int \frac{\mathrm{dx}}{{x}^{2} - x + 1}$

=$\int \frac{\left({x}^{2} - x + 1\right) \cdot \left(x + 1\right)}{{x}^{2} - x + 1} \cdot \mathrm{dx}$+$\int \frac{4 \mathrm{dx}}{4 {x}^{2} - 4 x + 4}$

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

=${x}^{2} / 2 + x + \frac{2 \sqrt{3}}{3} \arctan \left(\frac{2 x - 1}{\sqrt{3}}\right) + C$