Question #c5bb3

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Question #c5bb3

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Answer 1 · 2017-12-20T15:06:33

$3)$ $3)$ $\int \frac{{(arctan x)}^{2}}{1 + x^{2}} \cdot d x$ $int (arctanx)^2/(1+x^2)*dx$

= $\frac{{(arctan x)}^{3}}{3} + C$ $(arctanx)^3/3+C$

Explanation:

$3)$ $3)$ $\int \frac{{(arctan x)}^{2}}{1 + x^{2}} \cdot d x$ $int (arctanx)^2/(1+x^2)*dx$

After using $u = arctan x$ $u=arctanx$ , $x = tan u$ $x=tanu$ and $d x = {(sec u)}^{2} \cdot d u$ $dx=(secu)^2*du$ transforms, this integral became,

$\int \frac{u^{2} \cdot {(sec u)}^{2} \cdot d u}{{(sec u)}^{2}}$ $int (u^2*(secu)^2*du)/(secu)^2$

= $u^{2} \cdot d u$ $u^2*du$

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

= $\frac{{(arctan x)}^{3}}{3} + C$ $(arctanx)^3/3+C$

Answer 2 · 2017-12-20T15:38:51

$1)$ $1)$ $\int (x^{2} - 3 x) \cdot {(cos x)}^{2} \cdot d x$ $int (x^2-3x)*(cosx)^2*dx$

= $\frac{x^{3}}{6} - \frac{3 x^{2}}{4} + \frac{2 x^{2} - 6 x - 1}{8} \cdot sin 2 x + \frac{2 x - 3}{8} \cdot cos 2 x + C$ $x^3/6-(3x^2)/4+(2x^2-6x-1)/8*sin2x+(2x-3)/8*cos2x+C$

Explanation:

$1)$ $1)$ $\int (x^{2} - 3 x) \cdot {(cos x)}^{2} \cdot d x$ $int (x^2-3x)*(cosx)^2*dx$

= $\frac{1}{2} \int (x^{2} - 3 x) \cdot (1 + cos 2 x) \cdot d x$ $1/2int (x^2-3x)*(1+cos2x)*dx$

= $\frac{1}{2} \int (x^{2} - 3 x) \cdot d x$ $1/2int (x^2-3x)*dx$ + $\frac{1}{2} \int (x^{2} - 3 x) \cdot cos 2 x \cdot d x$ $1/2int (x^2-3x)*cos2x*dx$

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

$B = \int (x^{2} - 3 x) \cdot cos 2 x$ $B=int (x^2-3x)*cos2x$

= $(x^{2} - 3 x) \cdot \frac{1}{2} sin 2 x - (2 x - 3) (- \frac{1}{4} cos 2 x) + 2 (- \frac{1}{8} sin 2 x)$ $(x^2-3x)*1/2sin2x-(2x-3)(-1/4cos2x)+2(-1/8sin2x)$

= $\frac{2 x^{2} - 6 x - 1}{4} \cdot sin 2 x + \frac{2 x - 3}{4} \cdot cos 2 x$ $(2x^2-6x-1)/4*sin2x+(2x-3)/4*cos2x$

Thus,

$\int (x^{2} - 3 x) \cdot {(cos x)}^{2} \cdot d x$ $int (x^2-3x)*(cosx)^2*dx$

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

= $\frac{x^{3}}{6} - \frac{3 x^{2}}{4} + \frac{2 x^{2} - 6 x - 1}{8} \cdot sin 2 x + \frac{2 x - 3}{8} \cdot cos 2 x + C$ $x^3/6-(3x^2)/4+(2x^2-6x-1)/8*sin2x+(2x-3)/8*cos2x+C$

Answer 3 · 2017-12-20T15:46:43

$\int \frac{x^{3} + 2}{x^{2} - x + 1} \cdot d x$ $int (x^3+2)/(x^2-x+1)*dx$

= $\frac{x^{2}}{2} + x + \frac{2 \sqrt{3}}{3} arctan (\frac{2 x - 1}{\sqrt{3}}) + C$ $x^2/2+x+(2sqrt3)/3arctan((2x-1)/sqrt3)+C$

Explanation:

$2)$ $2)$ $\int \frac{x^{3} + 2}{x^{2} - x + 1} \cdot d x$ $int (x^3+2)/(x^2-x+1)*dx$

= $\int \frac{x^{3} + 1}{x^{2} - x + 1} \cdot d x$ $int (x^3+1)/(x^2-x+1)*dx$ + $\int \frac{d x}{x^{2} - x + 1}$ $int (dx)/(x^2-x+1)$

= $\int \frac{(x^{2} - x + 1) \cdot (x + 1)}{x^{2} - x + 1} \cdot d x$ $int ((x^2-x+1)*(x+1))/(x^2-x+1)*dx$ + $\int \frac{4 d x}{4 x^{2} - 4 x + 4}$ $int (4dx)/(4x^2-4x+4)$

= $\int (x + 1) \cdot d x$ $int (x+1)*dx$ + $2 \int \frac{2 d x}{{(2 x - 1)}^{2} + 3}$ $2int (2dx)/((2x-1)^2+3)$

= $\frac{x^{2}}{2} + x + \frac{2 \sqrt{3}}{3} arctan (\frac{2 x - 1}{\sqrt{3}}) + C$ $x^2/2+x+(2sqrt3)/3arctan((2x-1)/sqrt3)+C$

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