∫1+x43xdx is equal to

∫1+x43xdx Let x=t4dx=4t3dt=∫1+t3t24t3dt=4∫1+t3tdt Put 1+t=k3dt=3k2dk=3k2dk=4∫kk3−13k2dk=12∫k3k3−1dk=12∫k6−k3dk=12k77−k44+C=127k7−124k4+C=127(1+t)73−3(1+t)43+C=1271+x473−31+x443+C

∫1+x43xdx is equal to

∫1+x43xdx Let x=t4dx=4t3dt=∫1+t3t24t3dt=4∫1+t3tdt Put 1+t=k3dt=3k2dk=3k2dk=4∫kk3−13k2dk=12∫k3k3−1dk=12∫k6−k3dk=12k77−k44+C=127k7−124k4+C=127(1+t)73−3(1+t)43+C=1271+x473−31+x443+C

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$\int \frac{\sqrt[3]{1 + \sqrt[4]{x}}}{\sqrt{x}} d x is equal to$

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$3 {(1 + \sqrt[4]{x})}^{\frac{4}{3}} - \frac{12}{7} {(1 + \sqrt[4]{x})}^{\frac{7}{3}} + C$

$\frac{12}{7} {(1 + \sqrt[4]{x})}^{\frac{7}{3}} - 3 {(1 + \sqrt[4]{x})}^{\frac{4}{3}} + C$

$\frac{12}{7} {(1 + \sqrt[4]{x})}^{\frac{7}{3}} + 3 {(1 + \sqrt[4]{x})}^{\frac{4}{3}} + C$

$\frac{12}{7} {(1 + \sqrt[4]{x})}^{\frac{7}{3}} + \frac{1}{3} {(1 + \sqrt[4]{x})}^{\frac{4}{3}} + C$

answer is B.

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Detailed Solution

$\int \frac{\sqrt[3]{1 + \sqrt[4]{x}}}{\sqrt{x}} d x$

$Let x = t^{4}$

$\begin{array}{l} d x = 4 t^{3} d t \\ = \int \frac{\sqrt[3]{1 + t}}{t^{2}} 4 t^{3} d t \\ = 4 \int \sqrt[3]{1 + t} t d t \end{array}$

$Put 1 + t = k^{3}$

$\begin{array}{l} d t = 3 k^{2} d k \\ = 3 k^{2} d k \\ = 4 \int k (k^{3} - 1) 3 k^{2} d k \\ = 12 \int k^{3} (k^{3} - 1) d k \\ = 12 \int (k^{6} - k^{3}) d k \\ = 12 [\frac{k^{7}}{7} - \frac{k^{4}}{4}] + C \\ = \frac{12}{7} k^{7} - \frac{12}{4} k^{4} + C \\ = \frac{12}{7} (1 + t)^{\frac{7}{3}} - 3 (1 + t)^{\frac{4}{3}} + C \end{array}$