The value of \(\int \frac{x^{\frac{3}{2}}}{\sqrt{1+x^5}} d x\) is:

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  1. \(\frac{1}{2} \log \left(\sqrt{1+x^5}\right)+c\)
  2. \(\frac{2}{5} \log \left(x^{\frac{5}{2}}-\sqrt{1+x^5}\right)+c\)
  3. \(\frac{2}{5} \log \left(x^{\frac{5}{2}}+\sqrt{1+x^5}\right)+c\)
  4. \(\frac{1}{2} \log \left(\frac{1+x^5}{1-x^5}\right)+c\)

Answer (Detailed Solution Below)

Option 3 : \(\frac{2}{5} \log \left(x^{\frac{5}{2}}+\sqrt{1+x^5}\right)+c\)
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Detailed Solution

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Given:

\(\int \frac{x^{\frac{3}{2}}}{\sqrt{1+x^5}} d x\)

Concept:

Apply formula :

\(\rm \int\frac{1}{\sqrt{1+t^2}}dt =\log\left(t+\sqrt{1+t^2}\right)+C\)

Calculation:

\(\int \frac{x^{\frac{3}{2}}}{\sqrt{1+x^5}} d x\)

\(\rm=\int \frac{x^{\frac{3}{2}}}{\sqrt{1+(x^\frac{5}{2})^2}} d x\)

Substitute \(\rm x^{\frac{5}{2}}=t\implies x^{\frac{3}{2}}dx=\frac{2}{5}dt\)

Then

\(\rm =\frac{2}{5}\int\frac{1}{\sqrt{1+t^2}}dt\)

\(\rm =\frac{2}{5}\log\left(t+\sqrt{1+t^2}\right)+C\)

Undo substitution

\(\rm =\frac{2}{5}\log\left(x^{\frac{5}{2}}+\sqrt{1+x^5}\right)+C\)

Hence option (3) is correct.

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