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Question Number 103511 by bemath last updated on 15/Jul/20

$$\int \frac{dx}{\sqrt{x}(\sqrt[4]{x}+1)}=\mathrm{\_}\mathrm{\_}$$$$\left(a\right)-\frac{9\sqrt[4]{x}+1}{18{(\sqrt[4]{x}+1)}^9}+c$$$$\left(b\right)\frac{9\sqrt[4]{x}+1}{18{(\sqrt[4]{x}+1)}^9}+c$$$$\left(c\right)-\frac{9\sqrt[4]{x}-1}{18{(\sqrt[4]{x}+1)}^9}+c$$$$\left(d\right)\frac{9\sqrt[4]{x}+1}{8{(\sqrt[4]{x}+1)}^9}+c$$

Commented by bobhans last updated on 15/Jul/20

$$nothinganswer$$

Answered by bramlex last updated on 15/Jul/20

$$\mathrm{let}\mathrm{t}=1+\sqrt[4]{\mathrm{x}}\Rightarrow \sqrt[4]{\mathrm{x}}=\mathrm{t}-1$$$$\mathrm{x}={(\mathrm{t}-1)}^4\wedge \sqrt{\mathrm{x}}={(\mathrm{t}-1)}^2$$$$\mathrm{dx}=4{(\mathrm{t}-1)}^3\mathrm{dt}$$$$\mathrm{I}=\int \frac{4{(\mathrm{t}-1)}^3\mathrm{dt}}{{(\mathrm{t}-1)}^2.\mathrm{t}}=4\int \frac{\mathrm{t}-1}{\mathrm{t}}\mathrm{dt}$$$$=4\int (1-\frac{1}{\mathrm{t}})\mathrm{dt}$$$$=4\mathrm{t}-4\ln \mid \mathrm{t}\mid +\mathrm{c}$$$$=4(1+\sqrt[4]{\mathrm{x}})-4\ln \mid 1+\sqrt[4]{\mathrm{x}}\mid +\mathrm{c}$$

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