CUET Preparation Today
$\int \frac{2}{x^4-1} d x=$ |
$\log \left|\frac{x^2-1}{x^2+1}\right|+C$ $2 \tan ^{-1}\left(x^2\right)+C$ $\frac{1}{2} \log \left|\frac{\mathrm{x}-1}{\mathrm{x}+1}\right|-\tan ^{-1} \mathrm{x}+\mathrm{C}$ $\tan ^{-1} \mathrm{x}+\frac{1}{2} \log \left|\frac{\mathrm{x}-1}{\mathrm{x}+1}\right|+\mathrm{C}$ |
$\frac{1}{2} \log \left|\frac{\mathrm{x}-1}{\mathrm{x}+1}\right|-\tan ^{-1} \mathrm{x}+\mathrm{C}$ |
$\int(\frac{1}{x^2-1}-\frac{1}{x^2+1})dx$ So, $\frac{1}{2}log|\frac{x-1}{x+1}|-tan^{-1}x+C$ So, option 3 is correct. |
