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  • explain solution RD Sharma class 12 chapter Indefinite Integrals exercise 18.9 question 40 maths

explain solution RD Sharma class 12 chapter Indefinite Integrals exercise 18.9 question 40 maths

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Answer: \frac{1}{3}(\log x+x)^{3}+c

Hint: Use substitution method to solve this integral.

Given:   \int\left(\frac{x+1}{x}\right)(\log x+x)^{2} d x

Solution:

        \text { Let } I=\int\left(\frac{x+1}{x}\right)(\log x+x)^{2} d x

        \begin{aligned} &=\int\left(\frac{x}{x}+\frac{1}{x}\right)(\log x+x)^{2} d x \\ &=\int\left(1+\frac{1}{x}\right)(\log x+x)^{2} d x \end{aligned}

        \operatorname{Put} \log x+x=t \Rightarrow\left(\frac{1}{x}+1\right) d x=d t

        \Rightarrow d x=\frac{1}{\left(1+\frac{1}{x}\right)} \text { dt then }

        I=\int\left(1+\frac{1}{x}\right) t^{2} \frac{1}{\left(1+\frac{1}{x}\right)} d t=\int t^{2} d t

        I=\left[\frac{t^{2+1}}{2+1}\right]+c \quad\left[\because \int x^{n} d x=\frac{x^{n+1}}{n+1}+c\right]

            =\frac{t^{3}}{3}+c=\frac{1}{3}(\log x+x)^{3}+c \quad[\because t=\log x+x]

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