Gujarat BoardEnglish MediumSTD 12 ScienceMathsIndefinite Integrals2 Marks
Question
Write a value of$\int\frac{(\tan^{-1}\text{x})^3}{1+\text{x}^2}\text{dx}$
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Answer
Let $\text{I}=\int\frac{(\tan^{-1}\text{x})^3}{1+\text{x}^2}\text{dx}$ Let $\tan^{-1}\text{x}=\text{t}$ $\frac{1}{1+\text{x}^2}\text{dx}=\text{dt}$ $\therefore\ \text{I}=\int\frac{\text{t}^3}{1}\text{dt}$ $=\frac{\text{t}^4}{4}+\text{C}$ $\text{I}=\frac{(\tan^{-1}\text{x})^4}{4}+\text{C}$
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