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Question Number 45164 by Tinkutara last updated on 09/Oct/18

Commented by maxmathsup by imad last updated on 09/Oct/18

$$let\phi \left(x\right)={\left(\frac{x+a}{x+b}\right)}^{\frac{1}{4}}={\left(\frac{x+b+a-b}{x+b}\right)}^{\frac{1}{4}}={(1+\frac{a-b}{x+b})}^{\frac{1}{4}}\Rightarrow$$$${\phi }^{{}^{\prime }}\left(x\right)=\frac{1}{4}(-\frac{a-b}{{(x+b)}^2}){\left(\frac{x+a}{x+b}\right)}^{\frac{1}{4}-1}=\frac{b-a}{4}{(x+b)}^{-2}\frac{{(x+a)}^{-\frac{3}{4}}}{{(x+b)}^{-\frac{3}{4}}}$$$$=\frac{b-a}{4}{(x+b)}^{-2+\frac{3}{4}}{(x+a)}^{-\frac{3}{4}}=\frac{b-a}{4}{(x+a)}^{-\frac{3}{4}}{(x+b)}^{-\frac{5}{4}}\Rightarrow$$$$\frac{4}{b-a}{\phi }^{{}^{\prime }}\left(x\right)={\left\{{(x+a)}^{-3}{(x+b)}^{-5}\right\}}^{\frac{1}{4}}\Rightarrow$$$$\int {\left\{{(x+a)}^{-3}{(x+b)}^{-5}\right\}}^{\frac{1}{4}}=\frac{4}{b-a}{\left(\frac{x+a}{x+b}\right)}^{\frac{1}{4}}+cthecorrectansweris\left(C\right).$$

Commented by Tinkutara last updated on 09/Oct/18

But Sir it is by option checking, isn't there a method without seeing options?

Commented by MJS last updated on 10/Oct/18

$$\mathrm{I}{\mathrm{don}}^{\prime }\mathrm{t}\mathrm{trust}\mathrm{in}\left(\mathrm{b}\right)\mathrm{and}\left(\mathrm{d}\right)\mathrm{because}\mathrm{why}\mathrm{should}$$$$\mathrm{there}\mathrm{be}\mathrm{a}\mathrm{change}\mathrm{of}\mathrm{sign}?$$$$\int t^{\frac{1}{4}}dt=\frac{4}{5}{t}^{\frac{5}{4}}$$$$\mathrm{so}\mathrm{by}\mathrm{instinct}\mathrm{it}\mathrm{should}\mathrm{be}\mathrm{the}\mathrm{one}\mathrm{with}\mathrm{the}4$$$$\Rightarrow {\mathrm{I}}^{\prime }\mathrm{d}\mathrm{vote}\mathrm{for}\left(\mathrm{c}\right)$$$$\frac{d}{dx}\left[\frac{4}{b-a}{\left(\frac{x+a}{x+b}\right)}^{\frac{1}{4}}\right]=\frac{4}{b-a}\times \frac{d}{dx}\left[{\left(\frac{x+a}{x+b}\right)}^{\frac{1}{4}}\right]=$$$$=\frac{4}{b-a}\times \frac{1}{4}{\left(\frac{x+a}{x+b}\right)}^{-\frac{3}{4}}\times \frac{1\times (x+b)-1\times (x+a)}{{(x+b)}^2}=$$$$=\frac{1}{b-a}{\left(\frac{x+a}{x+b}\right)}^{-\frac{3}{4}}\times \frac{b-a}{{(x+b)}^2}={\left(\frac{x+b}{x+a}\right)}^{\frac{3}{4}}\times \frac{1}{{(x+b)}^2}=$$$$=\frac{{(x+b)}^{\frac{3}{4}-2}}{{(x+a)}^{\frac{3}{4}}}=\frac{{(x+b)}^{-\frac{5}{4}}}{{(x+a)}^{\frac{3}{4}}}={(x+b)}^{-\frac{5}{4}}{(x+a)}^{-\frac{3}{4}}=$$$$={\left({(x+a)}^{-3}{(x+b)}^{-5}\right)}^{\frac{1}{4}}$$

Commented by math khazana by abdo last updated on 10/Oct/18

$$siryoucantrythechangement\frac{x+a}{x+b}=t^4...$$

Commented by MJS last updated on 10/Oct/18

$$\mathrm{thank}\mathrm{you}\mathrm{Sir}$$

Commented by Tinkutara last updated on 10/Oct/18

Thank you very much Sir! I got the answer. ��������

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