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$\int \frac{\mathrm{x}^{9 / 2}}{\sqrt{1+\mathrm{x}^{11}}} \mathrm{dx}$ is -
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2119 Upvotes
Verified Answer
The correct answer is:
$\frac{2}{11} \log \left(\mathrm{x}^{7 / 2}+\sqrt{\mathrm{x}^7+1}\right)+\mathrm{c}$
$$
\begin{aligned}
& \text { Let } \mathrm{x}^{11 / 2}=\mathrm{t} \Rightarrow \frac{11}{2} \mathrm{x}^{9 / 2} \mathrm{dx}=\mathrm{dt} \\
& \Rightarrow \mathrm{I}=\frac{2}{11} \int \frac{\mathrm{dt}}{\sqrt{1+\mathrm{t}^2}}=\frac{2}{11} \log \left(\mathrm{t}+\sqrt{1+\mathrm{t}^2}\right)+\mathrm{c} \\
& =\frac{2}{11} \log \left(\mathrm{x}^{11 / 2}+\sqrt{1+\mathrm{x}^{11}}\right)+\mathrm{c} .
\end{aligned}
$$
\begin{aligned}
& \text { Let } \mathrm{x}^{11 / 2}=\mathrm{t} \Rightarrow \frac{11}{2} \mathrm{x}^{9 / 2} \mathrm{dx}=\mathrm{dt} \\
& \Rightarrow \mathrm{I}=\frac{2}{11} \int \frac{\mathrm{dt}}{\sqrt{1+\mathrm{t}^2}}=\frac{2}{11} \log \left(\mathrm{t}+\sqrt{1+\mathrm{t}^2}\right)+\mathrm{c} \\
& =\frac{2}{11} \log \left(\mathrm{x}^{11 / 2}+\sqrt{1+\mathrm{x}^{11}}\right)+\mathrm{c} .
\end{aligned}
$$
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