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Question Number 113323 by AbhishekBasnet last updated on 12/Sep/20

Answered by bemath last updated on 12/Sep/20

$$\mathrm{let}\mathrm{a}+\mathrm{bx}=\mathrm{u},\mathrm{x}=\frac{\mathrm{u}-\mathrm{a}}{\mathrm{b}}$$$$\int \left(\frac{\mathrm{u}-\mathrm{a}}{\mathrm{b}}\right){\mathrm{u}}^3\left(\frac{\mathrm{du}}{\mathrm{b}}\right)=\frac{1}{{\mathrm{b}}^2}\int ({\mathrm{u}}^4-{\mathrm{au}}^3)\mathrm{du}$$$$=\frac{1}{{\mathrm{b}}^2}(\frac{{\mathrm{u}}^5}{5}-\frac{{\mathrm{au}}^4}{4})+\mathrm{c}$$$$=\frac{{\mathrm{u}}^4}{{20\mathrm{b}}^2}(4\mathrm{u}-5\mathrm{a})+\mathrm{c}$$$$=\frac{{(\mathrm{a}+\mathrm{bx})}^4}{{20\mathrm{b}}^2}(4\mathrm{bx}-\mathrm{a})+\mathrm{c}$$

Answered by bemath last updated on 12/Sep/20

$$\mathrm{by}\mathrm{parts}\{\begin{array}{l}\mathrm{u}=\mathrm{x}\to \mathrm{du}=\mathrm{dx}\\ \mathrm{v}=\int {(\mathrm{a}+\mathrm{bx})}^3\mathrm{dx}=\frac{1}{4\mathrm{b}}{(\mathrm{a}+\mathrm{bx})}^4\end{array}$$$$\mathrm{I}=\frac{\mathrm{x}{(\mathrm{a}+\mathrm{bx})}^4}{4\mathrm{b}}-\int \frac{1}{4\mathrm{b}}{(\mathrm{a}+\mathrm{bx})}^4\mathrm{dx}$$$$\mathrm{I}=\frac{\mathrm{x}{(\mathrm{a}+\mathrm{bx})}^4}{4\mathrm{b}}-\frac{{(\mathrm{a}+\mathrm{bx})}^5}{{20\mathrm{b}}^2}+\mathrm{c}$$

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