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

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

Answers (1)

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

Solution:  

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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