Gujarat BoardEnglish MediumSTD 12 ScienceMathsINTEGRALS4 Marks
Question
Integrate the function in exercise. $\text{x}\ \tan^{-1}\text{x dx}$
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Answer
Let $\text{I}=\int\text{x}\tan^{-1}\text{x dx}$ Taking $\tan^{-1}\text{x}$ as first function and x as second function and integrating by parts, we obtain. $\text{I}=\tan^{-1}\text{x}\int\text{x} \ \text{dx}-\int\Bigg\{\Big(\frac{\text{d}}{\text{dx}}\tan^{-1}\text{x}\Big)\int\text{x} \ \text{dx}\Bigg\}\text{dx}$ $=\tan^{-1}\text{x}\Big(\frac{\text{x}^2}{2}\Big)-\int\frac{1}{1+\text{x}^2}.\frac{\text{x}^2}{2}\text{dx}$ $=\frac{\text{x}^2\tan^{-1}\text{x}}{2}-\frac{1}{2}\int\frac{\text{x}^2}{1+\text{x}^2}\text{dx}$ $=\frac{\text{x}^2\tan^{-1}\text{x}}{2}-\frac{1}{2}\int\Bigg(\frac{\text{x}^2+1}{1+\text{x}^2}-\frac{1}{1-\text{x}^2}\Bigg)\text{dx}$ $=\frac{\text{x}^2\tan^{-1}\text{x}}{2}-\frac{1}{2}\int\Bigg(1-\frac{1}{1+\text{x}^2}\Bigg)\text{dx}$ $=\frac{\text{x}^2\tan^{-1}\text{x}}{2}-\frac{1}{2}(\text{x}-\tan^{-1}\text{x})+\text{C}$ $=\frac{\text{x}^2}{2}\tan^{-1}\text{x}-\frac{\text{x}}{2}+\frac{1}{2}\tan^{-1}\text{x}+\text{C}$
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