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Question Number 96076 by bobhans last updated on 29/May/20

$$\int 3\mathrm{x}.2^{\mathrm{x}}\mathrm{dx}?$$

Answered by i jagooll last updated on 29/May/20

$$\int \left(3\mathrm{x}\right)\mathrm{d}\left(\frac{2^{\mathrm{x}}}{\ln 2}\right)=$$$$\frac{3\mathrm{x}.2^{\mathrm{x}}}{\ln 2}-3\int \frac{2^{\mathrm{x}}}{\ln 2}\mathrm{dx}=$$$$\frac{3\mathrm{x}.2^{\mathrm{x}}}{\ln 2}-\frac{3.2^{\mathrm{x}}}{{\left(\ln 2\right)}^2}+\mathrm{c}$$

Answered by abdomathmax last updated on 29/May/20

$$\mathrm{I}=\int 3\mathrm{x}{2}^{\mathrm{x}}\mathrm{dx}\Rightarrow \mathrm{I}=3\int \mathrm{x}{\mathrm{e}}^{\mathrm{xln}2}\mathrm{dx}\mathrm{by}\mathrm{parts}\mathrm{u}=\mathrm{x}$$$$\mathrm{and}{\mathrm{v}}^{{}^{\prime }}={\mathrm{e}}^{\mathrm{cln}2}\Rightarrow \mathrm{I}=3\{\frac{\mathrm{x}}{\ln 2}{\mathrm{e}}^{\mathrm{xln}2}-\int \frac{1}{\ln 2}{\mathrm{e}}^{\mathrm{xln}2}\mathrm{dx}\}$$$$=\frac{3\mathrm{x}}{\ln 2}\times 2^{\mathrm{x}}-\frac{3.2^{\mathrm{x}}}{{\left(\ln 2\right)}^2}+\mathrm{C}$$

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