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Question Number 135932 by Engr_Jidda last updated on 17/Mar/21

$$Evaluate\left(1\right){\int }_0^1{\int }_0^x{\int }_0^y(3x^2+2y^2-3z^2)dxdydz$$$$\left(2\right)\int {(2x-2)}^{3}dx$$$$\left(3\right)\int \left(\frac{x-5}{x^2-10x+2}\right)dx$$

Answered by Olaf last updated on 17/Mar/21

$$1)$$$$\mathrm{\Omega }={\int }_0^1{\int }_0^x{\int }_0^y(3x^2+2y^2-3z^2)dxdydz$$$$\mathrm{\Omega }={\int }_0^1{\int }_0^x{[3x^{2}z+2y^{2}z-z^3]}_0^{y}dxdy$$$$\mathrm{\Omega }={\int }_0^1{\int }_0^x(3x^{2}y+y^3)dxdy$$$$\mathrm{\Omega }={\int }_0^1{[\frac{3}{2}{x}^2{y}^2+\frac{y^4}{4}]}_0^{x}dx$$$$\mathrm{\Omega }={\int }_0^1\frac{7}{4}{x}^{4}dx=\frac{7}{4}$$$$2)$$$$\int {(2x-2)}^{3}dx=\frac{1}{8}{(2x-2)}^4+\mathrm{C}=2{(x-1)}^4+\mathrm{C}$$$$3)$$$$\int \frac{x-5}{x^2-10x+2}dx=\frac{1}{2}\int \frac{d(x^2-10x+2)}{x^2-10x+2}=\frac{1}{2}\ln \mid x^2-10x+2\mid +\mathrm{C}$$

Answered by Ar Brandon last updated on 17/Mar/21

$$\left(3\right)$$$$\mathrm{x}-5=\lambda \frac{\mathrm{d}}{\mathrm{dx}}({\mathrm{x}}^2-10\mathrm{x}+2)+\mu =\lambda (2\mathrm{x}-10)+\mu ,\lambda =\frac{1}{2},\mu =0$$$$\int \frac{\mathrm{x}-5}{{\mathrm{x}}^2-10\mathrm{x}+2}\mathrm{dx}=\frac{1}{2}\int \frac{2\mathrm{x}-10}{{\mathrm{x}}^2-10\mathrm{x}+2}\mathrm{dx}=\frac{\ln \mid {\mathrm{x}}^2-10\mathrm{x}+2\mid }{2}+\mathcal{C}$$

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