MCQ
$\int\frac{\text{x}}{4+\text{x}^4}\text{ dx}$ is equal to:
- A$\frac{1}{4}\tan^{-1}\text{x}^2+\text{C}$
- ✓$\frac{1}{4}\tan^{-1}\Big(\frac{\text{x}^2}{2}\Big)$
- C$\frac{1}{2}\tan^{-1}\Big(\frac{\text{x}^2}{2}\Big)$
- Dnone of these.
✓
Answer
Correct option: B.
$\frac{1}{4}\tan^{-1}\Big(\frac{\text{x}^2}{2}\Big)$
$\text{I}=\int\frac{\text{x}}{4+\text{x}^4}\text{ dx}$
Put $\text{x}^2=\text{t}$
$2\text{xdx}=\text{dt}$
$\text{x dx}=\frac{\text{dt}}{2}$
$\text{I}=\int\frac{\frac{\text{dt}}{2}}{4+\text{t}^2}$
$\text{I}=\frac{1}{2}\tan^{-1}\Big(\frac{\text{t}}{2}\Big)+\text{C}$
$\text{I}=\frac{1}{2}\tan^{-1}\Big(\frac{\text{x}^2}{2}\Big)+\text{C}$
Put $\text{x}^2=\text{t}$
$2\text{xdx}=\text{dt}$
$\text{x dx}=\frac{\text{dt}}{2}$
$\text{I}=\int\frac{\frac{\text{dt}}{2}}{4+\text{t}^2}$
$\text{I}=\frac{1}{2}\tan^{-1}\Big(\frac{\text{t}}{2}\Big)+\text{C}$
$\text{I}=\frac{1}{2}\tan^{-1}\Big(\frac{\text{x}^2}{2}\Big)+\text{C}$
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