Предмет: Алгебра,
автор: Аноним
ПОМОГИТЕ ПОЖАЛУЙСТА 2 ИНТЕГРАЛА
Приложения:
Ответы
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0
(9)
![\int\limits^1_0 {\frac{x}{(x^2+1)^2}} \, dx = \frac{1}{2}*\int\limits^1_0 {\frac{1}{(x^2+1)^2}} \, d(x^2) =\frac{1}{2}*\int\limits^1_0 {\frac{1}{(x^2+1)^2}} \, d(x^2+1) =\\\\
=\frac{1}{2}*\int\limits^{1^2+1}_{0^2+1} {t^{-2}}} \, dt =\frac{1}{2}*\int\limits^{2}_{1} {t^{-2}}} \, dt =\frac{1}{2}*\frac{t^{-1}}{-1}|^{2}_{1}=-\frac{1}{2}*\frac{1}{t}|^2_1=\\\\=
-\frac{1}{2}*[\frac{1}{2}-\frac{1}{1}]=\frac{1}{4}](https://tex.z-dn.net/?f=+%5Cint%5Climits%5E1_0+%7B%5Cfrac%7Bx%7D%7B%28x%5E2%2B1%29%5E2%7D%7D+%5C%2C+dx+%3D+%5Cfrac%7B1%7D%7B2%7D%2A%5Cint%5Climits%5E1_0+%7B%5Cfrac%7B1%7D%7B%28x%5E2%2B1%29%5E2%7D%7D+%5C%2C+d%28x%5E2%29+%3D%5Cfrac%7B1%7D%7B2%7D%2A%5Cint%5Climits%5E1_0+%7B%5Cfrac%7B1%7D%7B%28x%5E2%2B1%29%5E2%7D%7D+%5C%2C+d%28x%5E2%2B1%29+%3D%5C%5C%5C%5C%0A%3D%5Cfrac%7B1%7D%7B2%7D%2A%5Cint%5Climits%5E%7B1%5E2%2B1%7D_%7B0%5E2%2B1%7D+%7Bt%5E%7B-2%7D%7D%7D+%5C%2C+dt+%3D%5Cfrac%7B1%7D%7B2%7D%2A%5Cint%5Climits%5E%7B2%7D_%7B1%7D+%7Bt%5E%7B-2%7D%7D%7D+%5C%2C+dt+%3D%5Cfrac%7B1%7D%7B2%7D%2A%5Cfrac%7Bt%5E%7B-1%7D%7D%7B-1%7D%7C%5E%7B2%7D_%7B1%7D%3D-%5Cfrac%7B1%7D%7B2%7D%2A%5Cfrac%7B1%7D%7Bt%7D%7C%5E2_1%3D%5C%5C%5C%5C%3D%0A-%5Cfrac%7B1%7D%7B2%7D%2A%5B%5Cfrac%7B1%7D%7B2%7D-%5Cfrac%7B1%7D%7B1%7D%5D%3D%5Cfrac%7B1%7D%7B4%7D "\int\limits^1_0 {\frac{x}{(x^2+1)^2}} \, dx = \frac{1}{2}*\int\limits^1_0 {\frac{1}{(x^2+1)^2}} \, d(x^2) =\frac{1}{2}*\int\limits^1_0 {\frac{1}{(x^2+1)^2}} \, d(x^2+1) =\\\\
=\frac{1}{2}*\int\limits^{1^2+1}_{0^2+1} {t^{-2}}} \, dt =\frac{1}{2}*\int\limits^{2}_{1} {t^{-2}}} \, dt =\frac{1}{2}*\frac{t^{-1}}{-1}|^{2}_{1}=-\frac{1}{2}*\frac{1}{t}|^2_1=\\\\=
-\frac{1}{2}*[\frac{1}{2}-\frac{1}{1}]=\frac{1}{4}")
(10)
![\int\limits^8_0 {\frac{1}{\sqrt{3x+1}}} \, dx = \frac{1}{3}*\int\limits^8_0 {\frac{1}{\sqrt{3x+1}}} \, d(3x+1) =\frac{1}{3}*\int\limits^{3*8+1}_{3*0+1} {\frac{1}{\sqrt{t}}} \, dt =\\\\
=\frac{1}{3}* \int\limits^{25}_{1} {t^{-\frac{1}{2}}} \, dt=\frac{1}{3}*\frac{t^{\frac{1}{2}}}{\frac{1}{2}}|^{25}_1=\frac{2}{3}*\sqrt{t}|^{25}_3=\\\\
=\frac{2}{3}*[\sqrt{25}-\sqrt{1}]=\frac{2}{3}*[5-1]=\frac{8}{3}](https://tex.z-dn.net/?f=+%5Cint%5Climits%5E8_0+%7B%5Cfrac%7B1%7D%7B%5Csqrt%7B3x%2B1%7D%7D%7D+%5C%2C+dx+%3D+%5Cfrac%7B1%7D%7B3%7D%2A%5Cint%5Climits%5E8_0+%7B%5Cfrac%7B1%7D%7B%5Csqrt%7B3x%2B1%7D%7D%7D+%5C%2C+d%283x%2B1%29+%3D%5Cfrac%7B1%7D%7B3%7D%2A%5Cint%5Climits%5E%7B3%2A8%2B1%7D_%7B3%2A0%2B1%7D+%7B%5Cfrac%7B1%7D%7B%5Csqrt%7Bt%7D%7D%7D+%5C%2C+dt+%3D%5C%5C%5C%5C%0A%3D%5Cfrac%7B1%7D%7B3%7D%2A+%5Cint%5Climits%5E%7B25%7D_%7B1%7D+%7Bt%5E%7B-%5Cfrac%7B1%7D%7B2%7D%7D%7D+%5C%2C+dt%3D%5Cfrac%7B1%7D%7B3%7D%2A%5Cfrac%7Bt%5E%7B%5Cfrac%7B1%7D%7B2%7D%7D%7D%7B%5Cfrac%7B1%7D%7B2%7D%7D%7C%5E%7B25%7D_1%3D%5Cfrac%7B2%7D%7B3%7D%2A%5Csqrt%7Bt%7D%7C%5E%7B25%7D_3%3D%5C%5C%5C%5C%0A%3D%5Cfrac%7B2%7D%7B3%7D%2A%5B%5Csqrt%7B25%7D-%5Csqrt%7B1%7D%5D%3D%5Cfrac%7B2%7D%7B3%7D%2A%5B5-1%5D%3D%5Cfrac%7B8%7D%7B3%7D "\int\limits^8_0 {\frac{1}{\sqrt{3x+1}}} \, dx = \frac{1}{3}*\int\limits^8_0 {\frac{1}{\sqrt{3x+1}}} \, d(3x+1) =\frac{1}{3}*\int\limits^{3*8+1}_{3*0+1} {\frac{1}{\sqrt{t}}} \, dt =\\\\
=\frac{1}{3}* \int\limits^{25}_{1} {t^{-\frac{1}{2}}} \, dt=\frac{1}{3}*\frac{t^{\frac{1}{2}}}{\frac{1}{2}}|^{25}_1=\frac{2}{3}*\sqrt{t}|^{25}_3=\\\\
=\frac{2}{3}*[\sqrt{25}-\sqrt{1}]=\frac{2}{3}*[5-1]=\frac{8}{3}")
(10)
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