MHT CET · Maths · Indefinite Integration
The integral \(\int \frac{2 x^3-1}{x^4+x} \mathrm{~d} x\) is equal to
- A \(\log \frac{\left|x^3+1\right|}{x^2}+\mathrm{c}\), (where c is a constant of integration)
- B \(\frac{1}{2} \log \frac{\left(x^3+1\right)^2}{\left|x^3\right|}+\mathrm{c}\); (where c is a constant of integration)
- C \(\log \left|\frac{x^3+1}{x}\right|+\mathrm{c}\), (where c is a constant of integration)
- D \(\frac{1}{2} \log \frac{\left|x^3+1\right|}{x^2}+\mathrm{c}\), (where c is a constant of integration)
Answer & Solution
Correct Answer
(C) \(\log \left|\frac{x^3+1}{x}\right|+\mathrm{c}\), (where c is a constant of integration)
Step-by-step Solution
Detailed explanation
Let \(I=\int \frac{2 x^3-1}{x^4+x} \mathrm{~d} x=\int \frac{\left(2 x-\frac{1}{x^2}\right)}{\left(x^2+\frac{1}{x}\right)} \mathrm{d} x\)
\(\ldots\).[Dividing \(\mathrm{N}^{\mathrm{r}}\) and \(\mathrm{D}^{\mathrm{r}}\) by \(x^2\) ]
Put \(x^2+\frac{1}{x}=\mathrm{t} \Rightarrow\left(2 x-\frac{1}{x^2}\right) \mathrm{d} x=\mathrm{dt}\)
\(\begin{aligned}
\therefore \quad I & =\int \frac{d t}{t} \\
& =\log |t|+c \\
& =\log \left|x^2+\frac{1}{x}\right|+c=\log \left|\frac{x^3+1}{x}\right|+c
\end{aligned}\)
\(\ldots\).[Dividing \(\mathrm{N}^{\mathrm{r}}\) and \(\mathrm{D}^{\mathrm{r}}\) by \(x^2\) ]
Put \(x^2+\frac{1}{x}=\mathrm{t} \Rightarrow\left(2 x-\frac{1}{x^2}\right) \mathrm{d} x=\mathrm{dt}\)
\(\begin{aligned}
\therefore \quad I & =\int \frac{d t}{t} \\
& =\log |t|+c \\
& =\log \left|x^2+\frac{1}{x}\right|+c=\log \left|\frac{x^3+1}{x}\right|+c
\end{aligned}\)
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