$\int_{- 1}^{1} \frac{x^{2}}{\sqrt{x^{2} + 1} + x + 1} d x$ $int_(-1)^1x^2/(sqrt(x^2+1)+x+1)dx$ = ?

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$\int_{- 1}^{1} \frac{x^{2}}{\sqrt{x^{2} + 1} + x + 1} d x$ $int_(-1)^1x^2/(sqrt(x^2+1)+x+1)dx$ = ?

Find $\int_{- 1}^{1} \frac{x^{2}}{\sqrt{x^{2} + 1} + x + 1} d x$ $int_(-1)^1x^2/(sqrt(x^2+1)+x+1)dx$

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Answer 1 · 2018-04-03T22:16:27

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

Explanation:

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

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

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

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

= $\frac{1}{2} \int_{- 1}^{1} (x^{2} + x - x \sqrt{x^{2} + 1}) \cdot d x$ $1/2int_(-1)^1 (x^2+x-xsqrt(x^2+1))*dx$

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

= $\frac{1}{3}$ $1/3$

Answer 2 · 2018-04-03T22:28:58

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

Explanation:

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

After using $x = tan y$ $x=tany$ and $d x = {(sec y)}^{2} \cdot d y$ $dx=(secy)^2*dy$ transforms, this integral became

$\int_{- \frac{π}{4}}^{\frac{π}{4}} \frac{{(tan y)}^{2}}{sec y + tan y + 1} \cdot {(sec y)}^{2} \cdot d y$ $int_(-pi/4)^(pi/4) (tany)^2/(secy+tany+1)*(secy)^2*dy$

= $\int_{- \frac{π}{4}}^{\frac{π}{4}} \frac{{(tan y)}^{2} \cdot (tan y + 1 - sec y) \cdot {(sec y)}^{2} \cdot d y}{{(tan y + 1)}^{2} - {(sec y)}^{2}}$ $int_(-pi/4)^(pi/4) ((tany)^2*(tany+1-secy)*(secy)^2*dy)/((tany+1)^2-(secy)^2)$

= $\int_{- \frac{π}{4}}^{\frac{π}{4}} \frac{{(tan y)}^{2} \cdot (tan y + 1 - sec y) \cdot {(sec y)}^{2} \cdot d y}{{(tan y)}^{2} + 1 + 2 tan y - {(sec y)}^{2}}$ $int_(-pi/4)^(pi/4) ((tany)^2*(tany+1-secy)*(secy)^2*dy)/((tany)^2+1+2tany-(secy)^2)$

= $\int_{- \frac{π}{4}}^{\frac{π}{4}} \frac{{(tan y)}^{2} \cdot (tan y + 1 - sec y) \cdot {(sec y)}^{2} \cdot d y}{{(sec y)}^{2} + 2 tan y - {(sec y)}^{2}}$ $int_(-pi/4)^(pi/4) ((tany)^2*(tany+1-secy)*(secy)^2*dy)/((secy)^2+2tany-(secy)^2)$

= $\int_{- \frac{π}{4}}^{\frac{π}{4}} \frac{{(tan y)}^{2} \cdot (tan y + 1 - sec y) \cdot {(sec y)}^{2} \cdot d y}{2 tan y}$ $int_(-pi/4)^(pi/4) ((tany)^2*(tany+1-secy)*(secy)^2*dy)/(2tany)$

= $\frac{1}{2} \int_{- \frac{π}{4}}^{\frac{π}{4}} tan y \cdot (tan y + 1 - sec y) \cdot {(sec y)}^{2} \cdot d y$ $1/2int_(-pi/4)^(pi/4) tany*(tany+1-secy)*(secy)^2*dy$

= $\frac{1}{2} \int_{- \frac{π}{4}}^{\frac{π}{4}} {(tan y)}^{2} \cdot {(sec y)}^{2} \cdot d y$ $1/2int_(-pi/4)^(pi/4) (tany)^2*(secy)^2*dy$ + $\frac{1}{2} \int_{- \frac{π}{4}}^{\frac{π}{4}} (tan y) \cdot {(sec y)}^{2} \cdot d y$ $1/2int_(-pi/4)^(pi/4) (tany)*(secy)^2*dy$ - $\frac{1}{2} \int_{- \frac{π}{4}}^{\frac{π}{4}} {(sec y)}^{3} \cdot tan y \cdot d y$ $1/2int_(-pi/4)^(pi/4) (secy)^3*tany*dy$

= ${[\frac{1}{6} {(tan y)}^{3} + \frac{1}{4} {(tan y)}^{2} - \frac{1}{6} {(sec y)}^{3}]}_{- \frac{π}{4}}^{3}$ $[1/6(tany)^3+1/4(tany)^2-1/6(secy)^3]_(-pi/4)^(pi/4)$

= $\frac{1}{3}$ $1/3$

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$\int_{- 1}^{1} \frac{x^{2}}{\sqrt{x^{2} + 1} + x + 1} d x$ $int_(-1)^1x^2/(sqrt(x^2+1)+x+1)dx$ = ?

Find $\int_{- 1}^{1} \frac{x^{2}}{\sqrt{x^{2} + 1} + x + 1} d x$ $int_(-1)^1x^2/(sqrt(x^2+1)+x+1)dx$

2 Answers

Explanation:

Explanation:

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Find int_(-1)^1x^2/(sqrt(x^2+1)+x+1)dx∫1−1x2√x2+1+x+1dxint_(-1)^1x^2/(sqrt(x^2+1)+x+1)dx

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Explanation:

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