Question Number 9113 by hmhjkk last updated on 19/Nov/16
$$\int_{\mathrm{0}} ^{\mathrm{1}} \mathrm{x}^{\mathrm{2}} /\mathrm{x}^{\mathrm{2}} +\mathrm{1}=\mathrm{0} \\ $$
Commented by sou1618 last updated on 19/Nov/16
![∫_0 ^1 (x^2 /(x^2 +1))dx=∫_0 ^1 1−(1/(x^2 +1))dx (i) ∫_0 ^1 1dx=[x]_0 ^1 =1 (ii) ∫_0 ^1 (1/(x^2 +1))dx →_(x=tanθ) ∫_0 ^(π/4) (1/(tan^2 θ+1))×((d/dθ)tanθ)dθ =∫_0 ^(π/4) (1/((((sin^2 θ+cos^2 θ)/(cos^2 θ)))))×(1/(cos^2 θ))dθ =∫_0 ^(π/4) cos^2 θ×(1/(cos^2 θ))dθ =[θ]_0 ^(π/4) =(π/4) ((i),(ii))⇒ ans=1−(1/4)π](https://www.tinkutara.com/question/Q9118.png)
$$\int_{\mathrm{0}} ^{\mathrm{1}} \frac{{x}^{\mathrm{2}} }{{x}^{\mathrm{2}} +\mathrm{1}}{dx}=\int_{\mathrm{0}} ^{\mathrm{1}} \mathrm{1}−\frac{\mathrm{1}}{{x}^{\mathrm{2}} +\mathrm{1}}{dx} \\ $$$$ \\ $$$$\left({i}\right)\:\int_{\mathrm{0}} ^{\mathrm{1}} \mathrm{1}{dx}=\left[{x}\right]_{\mathrm{0}} ^{\mathrm{1}} =\mathrm{1} \\ $$$$ \\ $$$$\left({ii}\right)\:\int_{\mathrm{0}} ^{\mathrm{1}} \frac{\mathrm{1}}{{x}^{\mathrm{2}} +\mathrm{1}}{dx}\:\underset{{x}={tan}\theta} {\rightarrow}\:\int_{\mathrm{0}} ^{\pi/\mathrm{4}} \frac{\mathrm{1}}{{tan}^{\mathrm{2}} \theta+\mathrm{1}}×\left(\frac{{d}}{{d}\theta}{tan}\theta\right){d}\theta \\ $$$$\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:=\int_{\mathrm{0}} ^{\pi/\mathrm{4}} \frac{\mathrm{1}}{\left(\frac{{sin}^{\mathrm{2}} \theta+{cos}^{\mathrm{2}} \theta}{{cos}^{\mathrm{2}} \theta}\right)}×\frac{\mathrm{1}}{{cos}^{\mathrm{2}} \theta}{d}\theta \\ $$$$\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:=\int_{\mathrm{0}} ^{\pi/\mathrm{4}} {cos}^{\mathrm{2}} \theta×\frac{\mathrm{1}}{{cos}^{\mathrm{2}} \theta}{d}\theta \\ $$$$\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:=\left[\theta\right]_{\mathrm{0}} ^{\pi/\mathrm{4}} \\ $$$$\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:=\frac{\pi}{\mathrm{4}} \\ $$$$\left(\left({i}\right),\left({ii}\right)\right)\Rightarrow \\ $$$${ans}=\mathrm{1}−\frac{\mathrm{1}}{\mathrm{4}}\pi \\ $$