KCET · Maths · Indefinite Integration
The value of \(\int \frac{x^{2} d x}{\sqrt{x^{6}+a^{6}}}\) is equal to
- A \(\log \left|x^{3}+\sqrt{x^{6}+a^{6}}\right|+C\)
- B \(\log \left|x^{3}-\sqrt{x^{6}+a^{6}}\right|+C\)
- C \(\frac{1}{3} \log \left|x^{3}+\sqrt{x^{6}+a^{6}}\right|+C\)
- D \(\frac{1}{3} \log \left|x^{3}-\sqrt{x^{6}+a^{6}}\right|+C\)
Answer & Solution
Correct Answer
(C) \(\frac{1}{3} \log \left|x^{3}+\sqrt{x^{6}+a^{6}}\right|+C\)
Step-by-step Solution
Detailed explanation
Let \(I=\int \frac{x^{2}}{\sqrt{x^{6}+a^{6}}} d x\)
Let \(\quad x^{3}=t\)
\(\begin{aligned}
\Rightarrow \quad 3 x^{2} d x &=d t \\
I &=\frac{1}{3} \int \frac{1}{\sqrt{t^{2}+\left(a^{3}\right)^{2}}} d t \\
&=\frac{1}{3} \log \left|t+\sqrt{t^{2}+a^{6}}\right|+C \\
&=\frac{1}{3} \log \left|x^{3}+\sqrt{x^{6}+a^{6}}\right|+C
\end{aligned}\)
Let \(\quad x^{3}=t\)
\(\begin{aligned}
\Rightarrow \quad 3 x^{2} d x &=d t \\
I &=\frac{1}{3} \int \frac{1}{\sqrt{t^{2}+\left(a^{3}\right)^{2}}} d t \\
&=\frac{1}{3} \log \left|t+\sqrt{t^{2}+a^{6}}\right|+C \\
&=\frac{1}{3} \log \left|x^{3}+\sqrt{x^{6}+a^{6}}\right|+C
\end{aligned}\)
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