TS EAMCET · Maths · Indefinite Integration
\(\int\left[\frac{x^4-x}{x^{20}}\right]^{1 / 4} d x=\)
- A \(\frac{4}{15}\left(\frac{\left(x^3-1\right)^5}{x^{15}}\right)^{1 / 4}+C\)
- B \(\frac{4}{15}\left(\frac{x^4+1}{x^4}\right)^{1 / 4}+C\)
- C \(\frac{\sqrt{x^4+x^2+1}}{x}+C\)
- D \(\frac{3}{4}\left(x^{4 / 3}+x^{1 / 3}\right)+C\)
Answer & Solution
Correct Answer
(A) \(\frac{4}{15}\left(\frac{\left(x^3-1\right)^5}{x^{15}}\right)^{1 / 4}+C\)
Step-by-step Solution
Detailed explanation
Let \(\begin{aligned} I & =\int\left(\frac{x^4-x}{x^{20}}\right)^{1 / 4} d x \\ I & =\int \frac{1}{x^4}\left(1-\frac{1}{x^3}\right)^{1 / 4} d x\end{aligned}\)…
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