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provide solution for RD Sharma maths class 12 chapter Indefinite Integrals exercise  18.9 question 70

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Answer: \frac{1}{2} \log |1+\sqrt{x}|+c

Hint: Use substitution method to solve this integral

Given: \int \frac{1}{\sqrt{x}+x} d x


        \begin{aligned} &\text { Let } I=\int \frac{1}{\sqrt{x}+x} d x \\ &\text { Put } \sqrt{x}=t \Rightarrow \frac{1}{2 \sqrt{x}} d x=d t \\ &\Rightarrow d x=2 \sqrt{x} \; d t \text { then } \end{aligned}

       \begin{aligned} &I=\int \frac{1}{t+t^{2}} 2 \sqrt{x} \; d t \\ &=\int \frac{1}{t(1+t)} 2 \sqrt{x} \; d t=2 \int \frac{1}{1+t} d t \end{aligned}                \left[\begin{array}{l} \because \sqrt{x}=t \\ \Rightarrow x=t^{2} \end{array}\right]

        \begin{aligned} &\text { Again put } 1+t=u \Rightarrow d t=d u \text { then }\\ &I=2 \int \frac{1}{u} d u=\frac{1}{2} \log |u|+c \end{aligned}\left[\because \int \frac{1}{x} d x=\log |x|+c\right]

            =\frac{1}{2} \log |t+1|+c \quad[\because u=1+t]

        =\frac{1}{2} \log |1+\sqrt{x}|+c \quad[\because t=\sqrt{x}]


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