Category: Integration

Question Number 168368 by SUPERMATH last updated on 09/Apr/22 Answered by qaz last updated on 09/Apr/22 $$\int \frac{{\mathrm{x}}^2}{\sqrt{1+\mathrm{x}+{\mathrm{x}}^2}}\mathrm{dx}=\int \sqrt{1+\mathrm{x}+{\mathrm{x}}^2}-\frac{1+\mathrm{x}}{\sqrt{1+\mathrm{x}+{\mathrm{x}}^2}}\mathrm{dx}$$$$=\mathrm{x}\sqrt{\mathrm{1}+\mathrm{x}+\mathrm{x}^{\mathrm{2}} }−\int\frac{\mathrm{x}+\mathrm{2x}^{\mathrm{2}} }{\:\mathrm{2}\sqrt{\mathrm{1}+\mathrm{x}+\mathrm{x}^{\mathrm{2}}…

Question Number 37297 by math khazana by abdo last updated on 11/Jun/18 $$calculate{\int }_C\frac{z}{z^2+1}dzwithC=\{z\in C/\mid z\mid =\frac{1}{2}\}$$ Commented by math khazana by abdo last updated…

Question Number 168366 by cortano1 last updated on 09/Apr/22 Answered by qaz last updated on 09/Apr/22 $${\int }_{-\pi /2}^{\pi /2}\frac{\cos \mathrm{x}}{1-\mathrm{x}+\sqrt{{\mathrm{x}}^2+1}}\mathrm{dx}={\int }_{-\pi /2}^{\pi /2}\frac{\cos \mathrm{x}}{1+\mathrm{x}+\sqrt{{\mathrm{x}}^2+1}}\mathrm{dx}$$$$=\frac{\mathrm{1}}{\mathrm{2}}\int_{−\pi/\mathrm{2}}…

Question Number 37290 by math khazana by abdo last updated on 11/Jun/18 $$findf\left(\theta \right)={\int }_{-\mathrm{\infty }}^{+\mathrm{\infty }}{e}^{-x^2}cos\left(cos\theta x\right)dx.$$ Terms of Service Privacy Policy Contact:…

Question Number 37288 by math khazana by abdo last updated on 11/Jun/18 $$calculatef\left(\alpha \right)={\int }_{-\mathrm{\infty }}^{+\mathrm{\infty }}\frac{cos\left(2x\right)}{1+{ax}^2}dxwitha>0$$$$2)findthevalueof{\int }_{-\mathrm{\infty }}^{+\mathrm{\infty }}\frac{cos\left(2x\right)}{1+3x^2}dx.$$ Commented by…