What is the definite integral of x^2/(x^2+1) from 1 to 0 ?

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Answer 1 · 2018-05-04T18:34:15

$\int_{1}^{0}$ $int_1^0$ $= \frac{π}{4} - 1 = - 0.2146018366$ $=pi/4-1=-0.2146018366$

Starting out with the integral,

$\int_{1}^{0}$ $int_1^0$ $\frac{x^{2}}{x^{2} + 1} d x$ $x^2/(x^2+1) dx$

We want to get rid of $x^{2}$ $x^2$ ,

$\int_{1}^{0}$ $int_1^0$ $(\frac{x^{2} + 1}{x^{2} + 1} - \frac{1}{x^{2} + 1}) d x$ $((x^2+1)/(x^2+1)-1/(x^2+1)) dx$

$\int_{1}^{0}$ $int_1^0$ $(1 - \frac{1}{x^{2} + 1}) d x$ $(1-1/(x^2+1)) dx$

$\Rightarrow \int$ $=> int_$ $1 d x$ $1 dx$ - $\int$ $int_$ $\frac{1}{x^{2} + 1} d x$ $1/(x^2+1)dx$

Which gives,

$x - arctan (x) + C$ $x-arctan(x)+C$

$\frac{π}{4} + (- x) ∣_{0}^{1} \Rightarrow \frac{π}{4} - 1 = - 0.2146018366$ $pi/4+(-x)|_0^1 => pi/4-1=-0.2146018366$

This was a kinda strange integral since it goes from 0 to 1. But, these are the calculations I got to.