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Question Number 86850 by john santu last updated on 01/Apr/20

$$\int \frac{\mathrm{dx}}{{\mathrm{x}}^2{\left(\sqrt[4]{({\mathrm{x}}^4+1)}\right)}^3}$$

Commented by john santu last updated on 01/Apr/20

Answered by MJS last updated on 01/Apr/20

$$\int \frac{dx}{x^2{(x^4+1)}^{3/4}}=$$$$[t=\frac{{(x^4+1)}^{1/4}}{x}\to dx=-x^2{(x^4+1)}^{3/4}dt]$$$$=-\int dt=-t=-\frac{\sqrt[4]{x^4+1}}{x}+C$$

Commented by john santu last updated on 01/Apr/20

$$\mathrm{super}\mathrm{easy}\mathrm{sir}.\mathrm{how}\mathrm{to}\mathrm{know}$$$$\mathrm{t}=\frac{{({\mathrm{x}}^4+1)}^{3/4}}{\mathrm{x}}?\mathrm{by}\mathrm{observe}?$$

Commented by MJS last updated on 01/Apr/20

$$\mathrm{trying}...$$$$\frac{d}{dx}\left[{(x^4+1)}^{1/4}\right]=\frac{x^3}{{(x^4+1)}^{3/4}}$$$$\mathrm{then}\mathrm{knowing}\mathrm{that}\frac{d}{dx}\left[\frac{u}{v}\right]==\frac{u^{\prime }v-{uv}^{\prime }}{v^2}$$$$\mathrm{and}\mathrm{somehow}\mathrm{feeling}\mathrm{thar}v\mathrm{might}\mathrm{be}x$$

Commented by john santu last updated on 01/Apr/20

$$\mathrm{waw}..\mathrm{amazing}\mathrm{sir}.\mathrm{good}\mathrm{feeling}$$

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