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Question Number 134289 by bramlexs22 last updated on 02/Mar/21

$$\mathrm{I}=\int \frac{{\mathrm{x}}^{\mathrm{n}}}{\sqrt{\mathrm{ax}+\mathrm{b}}}\mathrm{dx}$$$$\mathrm{H}=\int \frac{{\mathrm{x}}^4}{\sqrt{2\mathrm{x}+1}}\mathrm{dx}$$

Answered by EDWIN88 last updated on 02/Mar/21

$${\mathrm{I}}_{\mathrm{n}}=\frac{{2\mathrm{x}}^{\mathrm{n}}\sqrt{\mathrm{ax}+\mathrm{b}}}{(1+2\mathrm{n})\mathrm{a}}-\frac{2\mathrm{nb}}{(1+2\mathrm{n})\mathrm{a}}.{\mathrm{I}}_{\mathrm{n}-1}$$

Answered by EDWIN88 last updated on 02/Mar/21

$$\mathrm{H}=\int \frac{{\mathrm{x}}^4}{\sqrt{2\mathrm{x}+1}}\mathrm{dx}$$$$\mathrm{let}2\mathrm{x}+1={\mathrm{w}}^2\Rightarrow 2\mathrm{dx}=2\mathrm{w}\mathrm{dw}\wedge \mathrm{x}=\frac{{\mathrm{w}}^2-1}{2}$$$$\mathrm{H}=\frac{1}{16}\int \frac{{({\mathrm{w}}^2-1)}^4\mathrm{w}}{\mathrm{w}}\mathrm{dw}=\frac{1}{16}\int {({\mathrm{w}}^2-1)}^4\mathrm{dw}$$

Answered by mathmax by abdo last updated on 02/Mar/21

$$\mathrm{I}=\int \frac{{\mathrm{x}}^{\mathrm{n}}}{\sqrt{\mathrm{ax}+\mathrm{b}}}\mathrm{dx}\mathrm{we}\mathrm{do}\mathrm{the}\mathrm{changement}\sqrt{\mathrm{ax}+\mathrm{b}}=\mathrm{t}\Rightarrow \mathrm{ax}+\mathrm{b}={\mathrm{t}}^2\Rightarrow$$$$\mathrm{ax}={\mathrm{t}}^2-\mathrm{b}\Rightarrow \mathrm{x}=\frac{{\mathrm{t}}^2-\mathrm{b}}{\mathrm{a}}(\mathrm{we}\mathrm{suppose}\mathrm{a}\ne 0)\Rightarrow$$$$\mathrm{I}=\frac{1}{{\mathrm{a}}^{\mathrm{n}}}\int \frac{{({\mathrm{t}}^2-\mathrm{b})}^{\mathrm{n}}}{\mathrm{t}}\left(\frac{2\mathrm{t}}{\mathrm{a}}\right)\mathrm{dt}=\frac{2}{{\mathrm{a}}^{\mathrm{n}+1}}\int {({\mathrm{t}}^2-\mathrm{b})}^{\mathrm{n}}\mathrm{dt}$$$$=\frac{2}{{\mathrm{a}}^{\mathrm{n}+1}}\int \sum _{\mathrm{k}=0}^{\mathrm{n}}{\mathrm{C}}_{\mathrm{n}}^{\mathrm{k}}{\mathrm{t}}^{2\mathrm{k}}{(-\mathrm{b})}^{\mathrm{n}-\mathrm{k}}\mathrm{dt}$$$$=\frac{2}{{\mathrm{a}}^{\mathrm{n}+1}}\sum _{\mathrm{k}=0}^{\mathrm{n}}{(-\mathrm{b})}^{\mathrm{n}-\mathrm{k}}{\mathrm{C}}_{\mathrm{n}}^{\mathrm{k}}\frac{1}{2\mathrm{k}+1}{\mathrm{t}}^{2\mathrm{k}+1}+\mathrm{C}$$$$=\frac{2}{{\mathrm{a}}^{\mathrm{n}+1}}\sum _{\mathrm{k}=0}^{\mathrm{n}}\frac{{(-\mathrm{b})}^{\mathrm{n}-\mathrm{k}}{\mathrm{C}}_{\mathrm{n}}^{\mathrm{k}}}{2\mathrm{k}+1}{\left(\sqrt{\mathrm{ax}+\mathrm{b}}\right)}^{2\mathrm{k}+1}+\mathrm{C}$$

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