JEE Mains · Maths · STD 12 - 7.1 indefinite integral
The integral \(\int {\frac{{2{x^3} - 1}}{{{x^4} + x}}} \,dx\) is equal to (Here \(C\) is a constant of integration)
- A \(\frac{1}{2}\,{\log _e}\frac{{\left| {{x^3} + 1} \right|}}{{{x^2}}} + C\)
- B \(\frac{1}{2}\,{\log _e}\frac{{{{\left| {{x^3} + 1} \right|}^2}}}{{\left| {{x^3}} \right|}} + C\)
- C \(\,{\log _e}\left| {\frac{{\left| {{x^3} + 1} \right|}}{x}} \right| + C\)
- D \(\,{\log _e}\frac{{\left| {{x^3} + 1} \right|}}{{{x^2}}} + C\)
Answer & Solution
Correct Answer
(C) \(\,{\log _e}\left| {\frac{{\left| {{x^3} + 1} \right|}}{x}} \right| + C\)
Step-by-step Solution
Detailed explanation
\(\int \frac{2 x^{3}-1}{x^{4}+x} d x\) \(\int \frac{2 x-\frac{1}{x^{2}}}{x^{2}+\frac{1}{x}} d x\) \(x^{2}+\frac{1}{x}=t\) \(\left(2 x-\frac{1}{x^{2}}\right) d x=d t\) \(\int \frac{\mathrm{dt}}{\mathrm{t}}=\ell \mathrm{n}(\mathrm{t})+\mathrm{C}\)…
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