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Need solution for RD Sharma Maths Class 12 Chapter 18 Indefinite Integrals Excercise Very Short Answers Question 17

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Answer: \frac{\left(\tan ^{-1} x\right)^{4}}{4}+c

Hint: You must know about the integral rule of tangent function.

Given: \int \frac{\left ( \tan^{-1} x\right )^{3}}{1+x^{2}}dx


I=\int \frac{\left ( \tan^{-1} x\right )^{3}}{1+x^{2}}dx

Let \tan^{-1}x=t and differentiate both sides,

                        \frac{1}{1+x^{2}}dx=dt                                                                  \left [ \frac{d}{dt} \tan^{-1}x=\frac{1}{1+x^{2}}\right ]

                \begin{aligned} &I=\int\left(t^{3}\right) d t \\ &=\frac{t^{4}}{4}+c \\ &{\left[\int x^{n} d x=\frac{x^{n+1}}{n+1}+c\right]} \\ &=\frac{\left(\tan ^{-1} x\right)^{4}}{4}+c \end{aligned}

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